A thin–tall Boolean algebra which is isomorphic to each of its uncountable subalgebras

Topology and its Applications 158 (2011) 1503–1525

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Topology and its Applications www.elsevier.com/locate/topol

A thin–tall Boolean algebra which is isomorphic to each of its uncountable subalgebras Robert Bonnet a,∗,1 , Matatyahu Rubin b a b

Université de Savoie, Chambery, France Ben Gurion University, Beer Sheva, Israel

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 27 June 2010 Received in revised form 1 May 2011 Accepted 3 May 2011

Theorem A (♦ℵ1 ). There is a Boolean algebra B with the following properties: (1) B is thin–tall, and (2) B is downward-categorical.

MSC: 54G12 06E15

That is, every uncountable subalgebra of B is isomorphic to B.

Keywords: Ostaszewski space Thin–tall Scattered space Superatomic Boolean algebra Rigid Boolean algebra

The algebra B from Theorem A has some additional properties. For an ideal K of B, set cmpl B ( K ) := {a ∈ B | a · b = 0 for all b ∈ K }. We say that K is almost principal if K ∪ cmpl B ( K ) generates B. (3) B is rigid in the following sense. Suppose that I, J are ideals in B and f : B / I → B / J is a homomorphism with an uncountable range. Then there is an almost principal ideal K of B such that |cmpl( K )|  ℵ0 , I ∩ K ⊆ J ∩ K , and for every a ∈ K , f (a/ I ) = a/ J . (4) The Stone space of B is sub-Ostaszewski. Boolean-algebraically, this means that: if I is an uncountable ideal in B, then B / I has cardinality  ℵ0 . (5) Every uncountable subalgebra of B contains an uncountable ideal of B. (6) Every subset of B consisting of pairwise incomparable elements has cardinality  ℵ0 . (7) Every uncountable quotient of B has properties (1)–(6). Assuming ♦ℵ1 we also construct a Boolean algebra C such that: (1) C has properties (1) and (4)–(6) from Theorem A, and every uncountable quotient of C has properties (1) and (4)–(6). (2) C is rigid in the following stronger sense. Suppose that I, J are ideals in C and f : C / I → C / J is a homomorphism with an uncountable range. Then there is a principal ideal K of C such that |cmpl( K )|  ℵ0 , I ∩ K ⊆ J ∩ K , and for every a ∈ K , f (a/ I ) = a/ J .

© 2011 Elsevier B.V. All rights reserved.

* 1

Corresponding author. E-mail addresses: [email protected] (R. Bonnet), [email protected] (M. Rubin). Robert Bonnet was supported by the “Center for Advanced Studies in Mathematics” of Ben Gurion University.

0166-8641/$ – see front matter doi:10.1016/j.topol.2011.05.003

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Elsevier B.V. All rights reserved.

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1. Introduction Definition 1.1. A Boolean algebra B is said to be downward-categorical, if B is uncountable and every uncountable subalgebra of B is isomorphic to B. The algebra of finite and cofinite subsets of ω1 is downward-categorical. We recall the notion of a thin–tall Boolean algebra (BA). Let X be a Boolean space. That is, X is a 0-dimensional compact Hausdorff space. Denote by Isol( X ) the set of isolated points of X , and define D ( X ) := X \ Isol( X ). The Cantor–Bendixon derivatives  of X are defined as follows: D 0 ( X ) = X , for any ordinal α , D α +1 ( X ) = D ( D α ( X )), and if δ is a limit ordinal then D δ ( X ) = γ <δ D γ ( X ). We say that X is thin–tall if for every α < ω1 , |Isol( D α ( X ))| = ℵ0 and |Isol( D ω1 ( X ))| < ℵ0 . A Boolean algebra B is thin–tall if its Stone space Ult( B ) (of all ultrafilters of B) is thin–tall. This work is motivated by the question whether thin–tall downward-categorical Boolean algebras exist. We have the following answer. Theorem 1.2. (Proved in Theorems 2.7 and 3.1.) Assume ♦ℵ1 . There is a thin–tall Boolean algebra which is downward-categorical. The proof of Theorem 1.2 is divided into two main claims: Theorem 2.7 and Theorem 3.1. In Theorem 2.7 we construct a Boolean algebra which we call a “condensed Boolean algebra”. The construction assumes ♦ℵ1 . Condensed BA’s are by definition thin–tall, and in Theorem 3.1 we prove that condensed Boolean algebras are downward-categorical. Condensedness implies several other Boolean algebraic properties. To state these, we need some additional definitions. A topological space is always nonempty, and in a Boolean algebra B, always 0 B = 1 B . A Boolean space X is scattered, if for some ordinal α , Dα ( X ) is finite, and it is unitary, if for some ordinal α , Dα ( X ) is a singleton. We denote the member of this singleton by e X . The rank of X , rk( X ), is the first ordinal α such that Dα ( X ) is finite. If F ⊆ X is closed, then rk X ( F ) denotes the rank of F with its induced topology. A Boolean algebra B is superatomic, if its Stone space is scattered, and it is unitary, if its Stone space is unitary. Assume that B is superatomic. The rank of B, rk( B ), is the rank of Ult( B ). For a ∈ B and a subset C of B set C  a := {c ∈ C | c  a}. If a = 0, then B  a is a BA, and its rank is denoted by rk B (a). Also define, rk B (0 B ) := −1. If J ⊆ B is an ideal, then rk( J ) is defined to be the rank of the BA J ∪ {−a | a ∈ J }. Define I ( B ) := {a ∈ B | rk B (a) < rk( B )}. Then I ( B ) is an ideal in B, and if B is unitary, then I ( B ) is a maximal ideal. Let B be a BA and E ⊆ B. The complement ideal of the set E, cmpl B ( E ), is defined as {b ∈ B | for every e ∈ E , b · e = 0}. Let B be a unitary BA and I ⊆ B be an ideal. We say that I is a secluded ideal, if: (1) I is non-principal. (2) rk(cmpl( I )) = rk( B ). (Hence I ⊆ I ( B ).) (3) For every a ∈ I ( B ) there are b ∈ I and c ∈ cmpl( I ) such that a = b + c. Topologically, this means the following. Let U IB ⊆ Ult( B ) be the open set corresponding to I . That is, U IB = {x ∈ Ult( B ) | x ∩ I = ∅}. Then I is secluded iff cl(U IB ) \ U IB = {e Ult( B ) } and rk(Ult( B ) \ U IB ) = rk(Ult( B )). Condensed Boolean algebras are rigid in the following strong sense. Theorem 1.3. (Proved in Corollary 4.4.) Let B be a condensed Boolean algebra, I , J be ideals of B, and f : B / I → B / J be a homomorphism. Suppose that |Rng( f )| = ℵ1 . Then there is a countable secluded ideal K of B such that I ∩ cmpl( K ) ⊆ J ∩ cmpl( K ), and for every a ∈ cmpl( K ), f (a/ I ) = a/ J . The rigidity property of Theorem 1.3 is better understood when stated topologically. The following theorem is equivalent to Theorem 1.3. (However, it is not the exact topological translation of Theorem 1.3.) Let clop( X ) denote the Boolean algebra of clopen sets of X . Theorem 1.3∗ . Let X be a Boolean space such that clop( X ) is condensed. Then for every closed subset Y ⊆ X and a continuous function ϕ : Y → X : if |Rng(ϕ )| = ℵ1 , then there is a countable open set V in X such that: (1) The boundary of V is {e X }. (2) Y \ V ⊆ Rng(ϕ ) \ V . (3) For every x ∈ Y \ V , ϕ (x) = x. The following special cases of Theorem 1.3 are more easily understood. Corollary 1.4. (Proved in Theorem 4.3 and Corollary 4.5.) (a) Let B be a condensed Boolean algebra, J be an ideal of B, and f : B → B / J be a homomorphism. Suppose that |Rng( f )| = ℵ1 . Then there is a countable secluded ideal K of B such that for every a ∈ cmpl( K ), f (a) = a/ J .

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(b) Let B be a condensed Boolean algebra, I be an ideal of B, and f : B / I → B be a homomorphism. Suppose that |Rng( f )| = ℵ1 . Then there is a countable secluded ideal K of B such that I ⊆ K , and for every a ∈ cmpl( K ), f (a/ I ) = a. The conclusion of Corollary 1.4(a) implies the conclusion of Theorem 1.3. In fact, Theorem 1.3∗ is the topological translation of Corollary 1.4(a) and not of Theorem 1.3. The trivial proof of the above implication appears in the proof of Corollary 4.4. Condensed Boolean algebras are not retractive. (See Definition 4.6.) In fact, in a condensed Boolean algebra, only trivial ideals have a retract. This fact is stated formally in the following theorem. Theorem 1.5. (Proved in Corollary 4.7(a).) Let B be a condensed Boolean algebra, I be an ideal of B. Then the following are equivalent. (1) I has a retract. (2) Either B / I is countable, or there is a countable secluded ideal K of B such that I ⊆ K . Remark. If I is an ideal in a condensed BA B, then B / I is countable iff I is uncountable. (Corollary 3.8.) A Boolean algebra B is said to be quotient-categorical, if | B | = ℵ1 , and every uncountable quotient of B is isomorphic to B. A condensed BA is not quotient-categorical. In fact, B / I ∼ = B only if I is trivial. That is: Corollary 1.6. Let I be an ideal in a condensed BA B. Then B / I ∼ = B iff there is a countable secluded ideal K of B such that I ⊆ K . This corollary is concluded from Corollary 1.4 and Theorem 1.5. 1.1. Sub-Ostaszewski and Ostaszewski algebras Definition 1.7. (a) A Boolean algebra B is a sub-Ostaszewski algebra, if Ult( B ) is thin–tall, and every closed subset of Ult( B ) is either countable or co-countable. (b) A Boolean algebra B is an Ostaszewski algebra if Ult( B ) is thin–tall and unitary, and for every closed subset F ⊆ Ult( B ): if e Ult( B ) is an accumulation point of F , then F is co-countable. Note that for superatomic BA’s the following are equivalent. (1) Every closed subset of Ult( B ) is either countable or co-countable. (2) For every uncountable ideal I ⊆ B, | B / I |  ℵ0 . Definition 1.7 is a translation of the topological notions of an Ostaszewski space and a sub-Ostaszewski space to the setting of Boolean algebras and Boolean spaces. The relevant topological definitions can be found in [6]. It follows trivially from the definitions that an Ostaszewski algebra is a sub-Ostaszewski algebra. A condensed BA is sub-Ostaszewski (Corollary 4.12). Indeed, the following holds. Proposition 1.8. (Proved in Lemma 3.4 and Corollary 4.12.) Let B be a condensed algebra and I ⊆ B be an uncountable ideal. Then there is a countable secluded ideal J such that I ⊇ cmpl( J ). Hence B is sub-Ostaszewski. However, a condensed algebra is never an Ostaszewski algebra. Every condensed BA is downward-categorical, and we shall next observe that: Observation 1.9. (Proved in Corollary 4.7(c).) A downward-categorical algebra is not an Ostaszewski algebra. A downward-categorical thin–tall BA must have countable secluded ideals. To see this, let B be a downward-categorical thin–tall BA. Let a ∈ B be such that B  a is countably infinite and unitary. Let A be the subalgebra of B generated by I ( B  a) ∪ I ( B  −a). Since A is uncountable, B ∼ = A. Also, I ( B  a) is a countable secluded ideal in A. So B has a countable secluded ideal. Let B be a thin–tall BA, I ⊆ B be a countable secluded ideal. Then e Ult( B ) is an accumulation point of U IB and

clUlt( B ) (U IB ) = U IB ∪ {e Ult( B ) }. Hence |clUlt( B ) (U IB )| = ℵ0 . It follows that B is not an Ostaszewski BA. In [9] (1996) Judith Roitman writes: “under ♦ there is an Ostaszewski space with all three properties”, where the three properties are: “(a) retractive; (b) isomorphic to every uncountable image; (c) isomorphic to every uncountable subalgebra”. Apparently, Theorem A of [9] is where the above statement is proved. However, Theorem A seems to prove only a restricted

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version of downward-categoricity. Let B be a thin–tall BA and A be a subalgebra of B. A is a bounded subalgebra of B, if for every b ∈ I ( B ) there is a ∈ A ∩ I ( B ) such that for every c ∈ A  b, c  a. The downward-categoricity statement implied by Theorem A seems to be:

• Every uncountable bounded subalgebra B is isomorphic to B. The following question remains open. Question 1.10. Is it consistent that there is a thin–tall downward-categorical and quotient-categorical Boolean algebra? We return to the rigidity property of condensed algebras (Theorem 1.3). A simplification of the construction of a condensed BA yields an Ostaszewski algebra which has a rigidity property slightly stronger than that of Theorem 1.3. Theorem 1.11. (Proved in Theorem 5.2, Proposition 5.3(b) and Corollary 5.4.) (a) (♦ℵ1 ) There is an Ostaszewski algebra B with the following property. Let I , J be ideals of B, and f : B / I → B / J be a homomorphism. Suppose that |Rng( f )| = ℵ1 . Then there is b ∈ I ( B ) such that I  −b ⊆ J  −b and for every a ∈ I ( B )  −b, f (a/ I ) = a/ J . (b) (The topological translation of Part (a).) There is an Ostaszewski space X such that for every closed subset Y ⊆ X and a continuous function ϕ : Y → X : if |Rng(ϕ )| = ℵ1 , then there is a countable clopen set V in X such that: (1) Y \ V ⊆ Rng(ϕ ) \ V . (2) For every x ∈ Y \ V , ϕ (x) = x. Theorem 1.11 is a restatement of Corollary 5.4 which deals with so-called “packed Boolean algebras”. We observe that a thin–tall downward-categorical BA cannot have the property of Theorem 1.11. Let B be such an algebra and I ⊆ B be a countable secluded ideal. Regard I as a Boolean ring. Then I has an automorphism f such that for every atom a of I , f (a) = a. Then f ∪ (Id  cmpl( I )) extends to an automorphism g of B, and there is no b ∈ I ( B ) such that g  −b = Id. The results of this work call for the following question. Question 1.12. Is it consistent that there is a thin–tall downward-categorical BA which is not sub-Ostaszewski? Theorem 1.2 was found by the authors in 1991. Judith Roitman [10] (2002) proved that the following is consistent: There exists an almost disjoint algebra which is downward-categorical. (An almost disjoint algebra is a subalgebra of P (ω) which is generated by the finite subsets of ω and an uncountable almost disjoint family.) The actual result of Roitman is stronger and deals with weak subalgebras. The analogous question about thin–tall quotient-categorical algebras was investigated by Bonnet and Shelah in [4], and by Roitman in [8] and [9]. Indeed, Bonnet and Shelah [4] constructed with the aid of ♦ℵ1 a thin–tall retractive quotientcategorical Boolean algebra. Roitman [8] constructed with the aid of CH a thin–tall quotient-categorical algebra. M. Weese (see Monk and Weese [1991]) proved: If B is downward-categorical, then B is superatomic, and if 2ℵ0 < 2ℵ1 holds, then B is thin–tall. Rigidity theorems for thin–tall Boolean algebras were obtained by M. Weese in [12] and by A. Dow and P. Simon in [5]. Weese, assuming (CH), proved that there is a thin–tall BA B such that for every f ∈ Aut( B ) there is α < ω1 such that for every a ∈ B, a/ I α ( B ) = f (a)/ I α ( B ). The same rigidity result was obtained by Dow and Simon assuming only ZFC. 2. The construction of a condensed Boolean algebra The construction of a condensed Boolean algebra is similar to the construction of a strongly concentrated BA in [11].  ( B ) := {a ∈ I ( B ) | B  a is unitary} and for every α < rk( B ) Let B be a superatomic BA. Set I α ( B ) := {a ∈ B | rk B (a) < α }, At  ( B ) | rk(a) = α }. Note that if α  rk( B ), then I α ( B ) is an ideal in B.  α ( B ) := {a ∈ At let At Let a, b ∈ B. Then a ∼ B b means that either (i) a, b = 0 B and rk B (a  b) < rk B (a), rk B (b), or (ii) a = b = 0 B . If A is a subset of B, set − A = {−a | a ∈ A }. An ideal I means a proper ideal. That is, 1 B ∈ / I . Note that I ∪ − I is a subalgebra of B and the isomorphism type of I ∪ − I does not depend on B. Recall that rk( I ) := rk( I ∪ − I ). For a member b ∈ B and a subset A ⊆ B set b · A := {b · a | a ∈ A }. Suppose that B is a subalgebra of C and c , d ∈ C . Then [c , d] B := {b ∈ B | c  b  d}. The intervals (c , d) B , (c , d] B and [c , d) B are defined in a similar way. Definition 2.1. Let B be a unitary BA, and I be an ideal in B. I is a pure ideal of B, if I is secluded and rk B (a) < rk( I ) for all a ∈ I.

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The verification of the next proposition is trivial and is left to the reader. Proposition 2.2. Let B be a unitary BA and I be an ideal in B. (a) Conditions (1), (2) and (3) below are equivalent. (1) I is secluded. B B Ult( B ) Ult( B ) }, and rk(U cmpl }) = rk(Ult( B )). (2) The boundary of both U IB and U cmpl ( I ) is {e ( I ) ∪ {e (3) Conditions (S1)–(S3) below hold. (S1) I is non-principal.  ( B ) there is a ∼ a such that either a ∈ I or a · I = {0}. (S2) For every a ∈ At  ( B ) such that for every s ∈ S, rk(s)  β (S3) For every β < rk( B ) there is an infinite set S of pairwise disjoint elements of At and s · I = {0}. Also, (S2) is equivalent to the following property: For every a ∈ I ( B ), there are a1 ∈ I and a2 ∈ cmpl B ( I ) such that a = a1 + a2 . (b) If I is secluded, then I ⊆ I ( B ). (c) If I is secluded, then there is no b ∈ I ( B ) such that I ⊆ B  b. (d) Let I be a secluded ideal. Then there are a pure ideal J 1 and a principal ideal J 2 such that J 1 ∩ J 2 = {0}, J 1 ∪ J 2 generates I , and either rk( J 2 ) = rk( I ), or J 2 = {0}. (e) Let I 1 be a secluded ideal or a principal ideal in B, and I 2 be a pure ideal in B. Assume that rk( I 1 ) < rk( I 2 ). Then the ideal I generated by I 1 ∪ I 2 is a pure ideal and rk( I ) = rk( I 2 ). Let B ∗ be a Boolean algebra, B ⊆ B ∗ be a subalgebra and d ∈ B ∗ . Then B  d is an ideal of B. The next definition introduces the notion of a PI-system. Essentially, a PI-system is a pair  B , I , where B is a unitary BA which is either countable or thin– tall, and I is a family of pure ideals of B. However, it is convenient to capture this situation by introducing another Boolean algebra B ∗ containing B and a set D ⊆ B ∗ \ B such that the family I of pure ideals is just { B  d | d ∈ D }. Definition 2.3. A pure ideal system (PI-system) is an object of the form M =  B  , B , D  where: (P1) (P2) (P3) (P4)

B  is an atomic Boolean algebra, B is a subalgebra of B  , B is unitary, D ⊆ B  \ B, D is infinite, and B ∪ D generates B  . At( B ) = At( B  ), and |At( B )| = ℵ0 . For every d ∈ D, B  d is a pure ideal in B. For every distinct d1 , d2 ∈ D, d1 · d2 ∈ B.

B  , B and D are denoted respectively by B M , B M and D M . Define rk( M ) to be rk( B ). Let I M denote the ideal of B  generated 

/ I M .) by I ( B ) ∪ D. (We shall soon verify in Proposition 2.4(b) that 1 B ∈

Proposition 2.4. Let M =  B  , B , D  be a PI-system. (a) (b) (c) (d)

For every b ∈ I ( B ), B   b = B  b. I M is a proper ideal of B M , and hence I M is a maximal ideal of B M . For every d ∈ D, B  d = I ( B )  d, and I ( B )  d is a maximal ideal in the algebra B   d. Let c ∈ B  and b ∈ B. If b · ( B  c ) = {0}, then b · c = 0.

 ( B ) be of minimal rank such that Proof. (a) Suppose by contradiction that for some a ∈ I ( B ), B   a = B  a. Let a ∈ At B   a = B  a. There is d ∈ D such that d · a ∈ / B. Since B  d is secluded there is a ∼ a such that either a  d or a · d = 0. By the minimality of a, d · (a − a ) ∈ B and d · (a − a) ∈ B. Suppose that a  d. Then









d · a = d · a − a − a + a − a



        = d · a − d · a − a + d · a − a = a − d · a − a + d · a − a ∈ B .

A contradiction. In the case that a · d = 0, we obtain that d · a = d · (a − a ) ∈ B. A contradiction.   (b) Suppose by contradiction that for some a ∈ I ( B ) and finite D 0 ⊆ D, a + D 0 = 1. Let e ∈ D \ D 0 . Then e = e · a + {e · d | d ∈ D 0 }. By (a) and (P4), e ∈ B. A contradiction. (c) Let d ∈ D. Suppose that b ∈ B  d. Recall that B  d is secluded. So it follows from Proposition 2.2(b) that b ∈ I ( B ). That is, B  d ⊆ I ( B )  d. Clearly, I ( B )  d is a proper ideal of B   d. So it suffices to show that I ( B )  d generates B   d. Trivially, {e · d | e ∈ D ∪ I ( B )} generates B   d. So it suffices to show that {e · d | e ∈ D ∪ I ( B )} ⊆ I ( B ). For every e ∈ D, e · d ∈ B. From the secludedness of B  d and Proposition 2.2(b) it follows that e · d ∈ I ( B ). Suppose next that, e ∈ I ( B ), then by (a), e · d ∈ I ( B ). (d) Suppose that b · c = 0. There is a ∈ At( B  ) such that a  b · c. Since At( B  ) = At( B ), it follows that a ∈ B. So a ∈ B  c. So 0 = a = b · a ∈ b · ( B  c ). 2

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The main property of a condensed BA B is that every uncountable subset of B is somewhere dense. We next define this notion. Definition 2.5. Let M =  B  , B , D  be a PI-system. (a) A wide interval of M is a set of the form [c , −e ] B , where c ∈ I ( B ), e ∈ I M and c · e = 0. We call c , e  a wide interval pair of M. Set Wip( M ) := {c , e  ∈ I ( B ) × I M | c · e = 0}. (b) A subset P ⊆ B is somewhere dense, if there is c , e  ∈ Wip( M ) such that for every c 1 , e 1  ∈ Wip( M ): if [c 1 , −e 1 ] B ⊆ [c , −e ] B , then P ∩ [c 1 , −e 1 ] B = ∅. Note that if c 1 , e 1 , c 2 , e 2  ∈ Wip( M ) and [c 1 , −e 1 ] B ⊆ [c 2 , −e 2 ] B , then c 2  c 1 and e 2  e 1 . Hence [c 1 , −e 1 ] B = [c 2 , −e 2 ] B iff c 1 = c 2 and e 1 = e 2 . The PI-system that we construct has the property that for every d ∈ D, B  d is countable. Since B is thin–tall, B  c is countable for every c ∈ I ( B ). So [c , −e ] B is uncountable for every c , e  ∈ Wip( M ). Definition 2.6. A PI-system M =  B  , B , D  satisfying (C1)–(C4) below is called a condensed PI-system. (C1) (C2) (C3) (C4)

B is a thin–tall BA. For every d ∈ D, rk( B  d) < ω1 . For every α < ω1 there is d ∈ D such that rk( B  d) > α . Every uncountable subset P of I ( B ) is somewhere dense.

A Boolean algebra B is called a condensed BA if B = B M for some condensed PI-system M. Theorem 2.7 (Construction Theorem). Assume ♦ℵ1 . There exists a condensed a PI-system. Remark 2.8. In the condensed PI-system M which we construct, the following holds. (C5) For every

α < ω1 , |{d ∈ D M | rk( B M  d)  α }|  ℵ0 .

This fact is not used when the various properties of condensed PI-systems are proved, except in Theorem 3.9(a). A condensed PI-system M fulfilling (C5) is called a narrow condensed PI-system, and B M is called a narrow condensed BA. The rest of this section concerns with the proof of Theorem 2.7. Definition 2.9. (a) Let B be a superatomic BA and A ⊆ B. Then A  B, means that for some α , A = I α ( B ) ∪ − I α ( B ). (b) A PI-system M is countable, if B M is countable. (c) Let M and N be PI-systems M  N, if B M  B N , D M ⊆ D N and for every d ∈ D M , B N  d = B M  d.  (d) Let { M i | i < α } be a sequence of PI-systems such that for every i < j < α , M i  M j . We define M = i <α M i to be

M :=



i <α



B Mi ,

i <α

B Mi ,





Di .

i <α

We shall construct an increasing a condensed PI-system.

ω1 -sequence of countable PI-systems { M i | i < ω1 } such that M =

 i <ω1

M i will be

Proposition 2.10.



(a) Let M be aPI-system and e ∈ I M . There are a finite subset D 0 of D and elements u , v ∈ I ( B ) such that u  D0 , v · and e = ( D 0 − u ) + v. In particular, if e ∈ I M , then e ∈ I ( B ) or B  e is a secluded ideal of B. (b) If M  N, then B M = B N ∩ B M . (c) If M  N, then D M = D N ∩ B M . (d) If M 1  M 2  M 3 , then M 1  M 3 . (e) Let { M i | i < α } and M be as in Definition 2.9(d) above. Then M is a PI-system, and for every i < α , M i  M.







D0 = 0

Proof. (a) Suppose that for i < k we have  e i = (di − u i )+   v i , where di ∈ D, and u i , v i ∈ I ( B ). Then i
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Let e ∈ I M . By the above paragraph it suffices to show that e is the sum of elements of the forms: (1) d − u, where d ∈ D and u ∈ I ( B ) and (2) v, where v ∈ I ( B ).   D 0 . So e = e · a + d∈ D 0 e · d. Now, e · a ∈ I ( B ). So it remains to For some finite subset D 0 ⊆ D and a ∈ I ( B ), e  a +  show that if e  d0 for some d0 ∈ D, then e has the desired form. Write e in disjunctive normal form as e =


j
dˆ j with c ∈ B and dˆ j ∈ D ∪ − D. We may deal with each e separately. Since e  d0 we may

assume that dˆ 0 = d0 . We may also assume that the dˆ j ’s are distinct. If c ∈ I ( B ), then e ∈ I ( B ). Suppose next that c ∈ − I ( B ). If for some j > 0, dˆ j ∈ D, then by (P4), dˆ 0 · dˆ j ∈ B. By Proposition 2.4(c), dˆ 0 · dˆ j ∈ I ( B ) and by Proposition 2.4(a), e ∈ I ( B ).

We are left with the case that c¯ := −c ∈ I ( B ) and for every j > 0, d¯ j := −dˆ j ∈ D. We have e = d0 − (¯c +



j >0 d0

· dˆ j ).

By (P4) and Proposition 2.4(a), d0 · dˆ j ∈ I ( B ). So e = d0 − u, where u ∈ I ( B ). (b) Suppose by contradiction that ( B N \ B M ) ∩ B M = ∅. Then ( B N \ B M ) ∩ I M = ∅. Let e ∈ ( B N \ B M ) ∩ I M . Then e ∈ I N .  / B M , D 0 = ∅. Write e in the form ( D 0 − u ) + v, where D 0 ⊆ D M and u , v ∈ I ( B M ). Since e ∈ If e ∈ − I ( B N ), then 1 = e + (−e ) ∈ I N . But this contradicts the properness of I N . So e ∈ I ( B N ). Hence e + u ∈ I ( B N ). By Proposition 2.4(a), B N  (e + u ) = B N  (e + u ). Let d ∈ D 0 . Then d ∈ D N and d  e + u. However, a member of D N cannot be  than a member I ( B N ). So we obtain a contradiction. I M , then 1 = d + (−d) ∈ I N , contradicting the properness (c) Suppose by contradiction that d ∈ ( D N \ D M ) ∩ B M . If d ∈ −    d u u v d = ( E − u ) + v, E ⊆ D , , v ∈ I ( B ) , < E and · E = 0. Hence · E = E0 − u ∈ / BM. of I N . So where 0 0 M M 0 0 0  E0 − u ∈ / B N . So for some e ∈ E 0 , d · e ∈ / B N . But d = e. So (P4) is contradicted. A contradiction. By (b), (d) and (e) are trivial. 2

Definition 2.11. (a) A subset Q ⊆ Wip( M ) is a dense set of intervals in M, if for every c , e  ∈ Wip( M ), there is c 1 , e 1  ∈ Q such that [c 1 , −e 1 ] ⊆ [c , −e ]. P := {c , e  ∈ Wip( M ) | P ∩ [c , −e ] = ∅}. We say that P ⊆ B M is nowhere dense in M, if Q MP is a dense (b) For P ⊆ B M let Q M set of intervals in M. Note that P is nowhere dense iff it is not somewhere dense. (c) Let M  N be PI-systems and Q ⊆ Wip( M ). N is convenient for  Q , M  (notation: Cnvnt( Q , M ; N )) if Q is a dense set of intervals in M, and for every a ∈ I ( B N ) \ B M there is c , e  ∈ Q such that a ∈ [c , −e ]. Note that for Q ⊆ Wip( M ), Q is a dense set of intervals in M iff Cnvnt( Q , M ; M ) holds. So for P ⊆ B M , P is nowhere P , M ; M ) holds. dense in M iff Cnvnt( Q M We construct { M i | i < ω1 } by transfinite induction. We may assume that in a PI-system M, B M is a subalgebra of P (ω) and At( B M ) = At( B M ) = {{i } | i ∈ ω}. The following lemma will be used in the construction of M α +1 from M α . Lemma 2.12 (Successor Case Lemma). (a) Let M be a countable PI-system. Let { Q i , M i  | i ∈ ω} be such that for every i ∈ ω : M i  M; either M i = M or rk( M i ) < rk( M ); and Cnvnt( Q i , M i ; M ) holds. Then there is a countable PI-system N such that M  N, rk( N ) = rk( M ) + 1, D N = D M , and for every i ∈ ω , Cnvnt( Q i , M i ; N ) holds. (b) For a BA B and a subset C ⊆ B we denote by cl B (C ) the subalgebra of B generated by C . If N is a countable PI-system, then there is d ⊆ ω such that N [d] := clP (ω) ( B N ∪ {d}), B N , D N ∪ {d} is a PI-system, d ∈ / B N and rk( B N  d) = rk( N ). (c) Let N and d be as in Part (b). Then: (1) N  N [d]. (2) For every  Q , M : if Cnvnt( Q , M ; N ) holds, then Cnvnt( Q , M ; N [d]) holds. Suppose that δ is a limit ordinal, and for every i < δ , M i has been constructed. M δ is defined to be following trivial observation assures that the induction hypotheses hold in the limit case.

 i <δ

M i . The

Proposition 2.13 (Limit Case Proposition). Let α be a countable limit ordinal and { N i | i < α } be a sequence of PI-systems, such that for every i < j < α , N i  N j . For every i < α , let { Q i , , M i ,  | ∈ L i } be such that for every j  i and ∈ L i , Cnvnt( Q i , , M i , ; N j ).  Then for every i < α and ∈ L i , Cnvnt( Q i , , M i , ; j <α N j ). Proof. The proof amounts to checking the definitions.

2

The central ideas in the proof of the existence of a thin–tall downward-categorical BA appear in the proofs of Theorem 2.7 (Construction Theorem), and Theorem 3.1. The main computational part of the proof appears in the Successor Case Lemma. We start with the computational part. To handle Part (a) of Lemma 2.12, we introduce the following notion of countable forcing.

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Definition 2.14. Let M =  B  , B , D  be a countable PI-system. p (a) We define the partially ordered set P = P M . A member p ∈ P has the form {c i | i ∈ ω}, σ p , where: p

p

(1) {c i | i ∈ ω} is a pairwise disjoint subset of I ( B ) such that { j ∈ ω | c j = ∅} is finite; (2)

σ p is a finite set of triples of the form i , d, b, where i ∈ ω, d ∈ D and b = c ip ∩ d. p

q

Let p , q ∈ P . Then p  q, if for every i ∈ ω , c i ⊆ c i , and

σ p ⊆ σ q.



p

(b) Let G be a directed subset of P . For every i ∈ ω let c iG = {c i | p ∈ G }. Let B GM be the subalgebra of P (ω) generated by B ∪ {c iG | i ∈ ω}, ( B M )G be the subalgebra of P (ω) generated by B GM ∪ D, and M G = ( B M )G , B GM , D . We next define some subsets of P . We shall prove that these subsets are dense in P , and that if G is a directed subset of P meeting all these sets, then M G is the N required in Part (a) of the “Successor Case Lemma”. In other words, we have a family of tasks. For each task we define a subset of P corresponding to this task. Fulfilling a task is achieved by making sure that G intersects the subset of P which corresponds to this task. Fulfilling all tasks will assure that M G is as required in Part (a) of the “Successor Lemma”.  Case p (c) For p ∈ P let c p = i ∈ω c i . For b ∈ I ( B ) define T b = T bM as follows:





T b := p ∈ P b ⊆ c p . Meeting T b will assure that b is contained in a finite union of c iG ’s, and that for every i ∈ ω , c iG ∩ b ∈ B. (d) Let i , k ∈ ω and β < rk( M ). Define T i ,k,β = T iM ,k,β as follows:



T i ,k,β = p ∈ P there is a pairwise disjoint subset S ⊆ I ( B )

p

such that |S | = k and for every a ∈ S : rk B (a)  β and a ⊆ c i . Meeting all the T b ’s and all T i ,k,β ’s will assure that B M  B GM and that rk( B GM ) = rk( B M ) + 1. (e) For i ∈ ω and d ∈ D, define T i ,d = T iM ,d as follows:









p T i ,d = p ∈ P i , d, c i ∩ d ∈ σ p .

Meeting all T b ’s, T i ,k,β ’s and T i ,d ’s assures that M  M G . = M . It is the set of all objects t of the form t = ηt , at , bt , where ηt is a nonempty (f) We define the set of terms t t finite subset of ω , and a , b are disjoint members  of I ( B ). For a directed subset G ⊆ P define t G := ( {c iG | i ∈ ηt } \ at ) ∪ bt . It will follow that if G meets all T b ’s and all T i ,k,β ’s, then {t G | t ∈ M } = I ( B GM ) \ I ( B M ). (g) For t ∈ and p ∈ P , we define the interval of t and p. It is denoted by intrvl(t , p ) and has the following properties: (i) It is a wide interval; (ii) It is the maximal interval I with the property that for every directed G ⊆ P containing p, t G ∈ I . To define intrvl(t , p ), we first define g (t , p ) and h(t , p ) . Once these are defined we set intrvl(t , p ) := [ g (t , p ) , −h(t , p ) ]. Define

τ τ

τ τ

g (t , p ) := p



For i ∈ ω let h i :=

h(t , p ) :=





p c i i ∈ ηt \ at ∪ bt .





p {d \ b | i , d, b ∈ σ p }. For ζ ⊆ ω let cˆ ζ :=







p

{c i | i ∈ ζ }. For t ∈ τ and p ∈ P , let

p p h i i ∈ ηt ∪ cˆ ω\ηt ∪ at \ bt .

Note that  g (t , p ) , h(t , p )  ∈ Wip( M ), and if G ⊆ P is directed and p ∈ G, then t G ∈ intrvl(t , p ). (h) Suppose L  M and Q ⊆ Wip( L ) fulfill Cnvnt( Q , L ; M ) and that either L = M or rk( L ) < rk( M ). For t ∈ T t , Q = T tM, Q as follows:



τM

define



T t , Q = p ∈ P there is c , e  ∈ Q such that intrvl(t , p ) ⊆ [c , −e ] . Meeting T t , Q will assure that t G lies in an interval belonging to Q . The proof of the next two lemmas is technical but straightforward. We skip their proof. Part (b) of the following lemma will serve as a convenient intermediate step in the proof of the lemma that follows it. Lemma 2.15 (Interval Lemma). Let M be a countable PI-system, t ∈

τ M and p ∈ P M . Set P = P M .

(a) Let G ⊆ P be directed and p ∈ G. Then t G ∈ intrvl(t , p ). If p  q, then intrvl(t , p ) ⊆ intrvl(t , q). (b) Let  g , h ∈ Wip( M ) be such that g (t , p ) ⊆ g and h(t , p ) ⊆ h. Then there is q  p such that intrvl(t , q) ⊆ [ g , −h].

R. Bonnet, M. Rubin / Topology and its Applications 158 (2011) 1503–1525

Lemma 2.16 (Density Lemma). Let M =  B  , B , D  be a countable PI-system, P = P M and (a) For every b ∈ I ( B ), T bM is dense in P . (b) For every i , k ∈ ω and β < rk( M ), T iM ,k,β is dense in P . is dense in P . (c) For every i ∈ ω and d ∈ D, T iM ,d

(d) Let L, Q be as in Definition 2.14(h). Then for every t ∈

τ , T tM, Q

1511

τ = τ M.

is dense in P .

Proof of Lemma 2.12 (Successor Case Lemma). (a) Let M and { Q i , M i  | i ∈ ω} be as in Lemma 2.12(a). Let G be a directed subset of P M such that G intersects all the dense subsets of P defined in Definition 2.14(c), (d), (e) and (h). That is, (1) G ∩ T bM = ∅ for every b ∈ I ( B M ); (2) G ∩ T iM ,k,β = ∅ for every i , k ∈ ω and β < rk( M ); (3) G ∩ T iM ,d = ∅ for every i ∈ ω and d ∈ D M ;

(4) G ∩ T tM, Q = ∅ for every t ∈ j

τM

and j ∈ ω .

Let N be defined as follows: B N := clP (ω) ( B M ∪ {c iG | i ∈ ω}), D N := D M and B N := clP (ω) ( B N ∪ D N ). We check that N is as required. At first we show that N is a PI-system. In fact, we show that if G intersects every dense set of the forms T bM , M T iM ,k,β and T i ,d , then N is a PI-system. The following facts imply that B N is superatomic: (1) B N is generated by B M ∪ {c iG | i ∈ ω}, (2) B M is superatomic, (3) {c iG | i ∈ ω} is a partition of ω . Since {{n} | n ∈ ω} ⊆ B N ⊆ B N ⊆ P (ω), it follows that B N is atomic and that At( B N ) = At( B N ) and |At( B N )| = ℵ0 . So N fulfills (P2). Since M is a PI-system, B M ⊆ B N and D N = D M , it follows that for every d1 , d2 ∈ D N , d1 ∩ d2 ∈ B N . So N fulfills (P4). p Fact 1: For every b ∈ I ( B M ), B N  b = B M  b. Let b ∈ I ( B M ). Choose p ∈ G ∩ T b . Then for every i ∈ ω , b ∩ c iG = c i ∩ b ∈ I ( B M ). Hence B N  b = B M  b. 2 Fact 2: For every i ∈ ω and a ∈ B N  c iG , either a ∈ I ( B M ) or c iG \ a ∈ I ( B M ). Let a ∈ B N  c iG . Then either (i) there is b ∈ I ( B M ) such that a = c iG ∩ b or (ii) there is b ∈ I ( B M ) such that a = c iG \ b. This is so since for every j = i, c Gj ∩ c iG = ∅ and since I ( B M ) is a maximal ideal of B M . By Fact 1, if (i) happens then a ∈ I ( B M ), and if (ii) happens, then c iG \ a ∈ I ( B M ). 2 Fact 3: For every i ∈ ω , rk( B M  c iG ) = rk( B M ). Since B M  c iG is an ideal in B M , rk( B M  c iG )  rk( B M ). Let i ∈ ω , β < rk( B M ) and k ∈ ω . Choose p ∈ G ∩ T i ,k,β . Let S ⊆ I ( B ) be a pairwise disjoint set such that |S | = k and for every p

a ∈ S : rk B M (a)  β and a ⊆ c i . So rk( B M  c iG ) > β . We conclude that rk( B M  c iG )  sup{β + 1 | β < rk( B M )} = rk( B M ). So rk( B M  c iG ) = rk( B M ). 2

Fact 4: For every i ∈ ω , c iG ∈ / B M . Suppose by contradiction that c iG ∈ B M . By Fact 3, rk B M (c iG ) = rk( B M  c iG ) = rk( B M ).

Let j = i. Then rk B M (−c iG ) = rk( B M  −c iG )  rk( B M  c Gj ) = rk( B M ). This implies that B M is not unitary. A contradiction. So c iG ∈ / BM.

2

Fact 5: For every i ∈ ω , rk B N (c iG ) = rk( B M ). By Facts 2 and 4, B N  c iG ∼ = ( B M  c iG ) ∪ −( B M  c iG ). So













rk B N c iG = rk B N  c iG = rk B M  c iG = rk( B M ).

2

Fact 6: rk( B N ) = rk( B M ) + 1 and B N is unitary. For every a ∈ B N there is a finite set

−a ⊆



BN

ρ ⊆ ω such that a ⊆

Since for every i ∈ ω , rk = rk( B M ), it follows that rk( B N ) = rk( B M ) + 1 and B N is unitary. i ∈ρ Clause (P1) in the definition of a PI-system consisted of the following items:

(P1.1) (P1.2) (P1.3) (P1.4)

c iG .

(c iG )

 2

G i ∈ρ c i

or

B  is an atomic Boolean algebra, B is a subalgebra of B  , B is unitary, D ⊆ B  \ B, D is infinite, B ∪ D generates B  .

We already know that N satisfies (P1.1), (P1.2) and (P1.4) and that D N is infinite. It remains to check that D N ⊆ B N \ B N . Fact 7: For every d ∈ D N and i ∈ ω , c iG ∩ d ∈ I ( B M ). Let d ∈ D N and i ∈ ω . Then d ∈ D M . Let p ∈ G ∩ T i ,d . So i , d, p

q

p

p

p

d ∩ c i  ∈ σ p . So for every q ∈ G: if q  p, then c i ∩ d = d ∩ c i . Hence c iG ∩ d = d ∩ c i ∈ I ( B M ). The fact that d ∩ c i ∈ I ( B M ) follows from Proposition 2.4(a). 2 Fact 8: Every e ∈ B N has the one of following forms:

  c G ) \ a) ∪ b ∪ (( i ∈ρ −c iG ) \ c ), where ρ , ζ are finite subsets of ω and a, b, c ∈ I ( B M ); i ∈ζ iG (II) (( i ∈ζ c i ) \ a) ∪ b, where ζ is a finite subset of ω and a, b ∈ I ( B M ). (I) ((

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R. Bonnet, M. Rubin / Topology and its Applications 158 (2011) 1503–1525

The proof of Fact 8 uses the following facts: (i) Fact 1, (ii) {c iG | i ∈ ω} is a partition of ω . The rest of the verification is left to the reader. 2  Fact 9: Let d ∈ D N . Then there are no ζ , b such that ζ is a finite subset of ω , b ∈ I ( B M ) and d ⊆ ( i ∈ζ c iG ) ∪ b. Suppose



otherwise. Then d = (d ∩ (

i ∈ζ

c iG )) ∪ (d ∩ b) ∈ I ( B M ) ⊆ B M . A contradiction.

2



Fact 10: Let d ∈ D N . Then there are no

ρ , c such that ρ is a finite subset of ω, c ∈ I ( B M ) and ( i∈ρ −c iG ) \ c ⊆ d. Suppose otherwise. Since G intersects T c , ξ := {i ∈ ω | c iG ∩ c = ∅} is finite. Let j ∈ ω \ (ρ ∪ ξ ). Then c Gj ⊆ d. So by Fact 7, c Gj ∈ I ( B M ) ⊆ B M . This contradicts Fact 4. 2 Fact 11: D N ⊆ B N \ B N . Suppose by contradiction that d ∈ D N ∩ B N . Then d has either form (I) or form (II) of Fact 8. If  the former happens, then ( i ∈ρ −c iG ) \ c ⊆ d, where ρ is a finite subset of ω and c ∈ I ( B M ). This contradicts Fact 10. If the



latter happens, then d ⊆ ( i ∈ζ c iG ) ∪ b, where ζ is a finite subset of ω and b ∈ I ( B M ). This contradicts Fact 9. 2 We have shown that N fulfills Clause (P1) in the definition of a PI-system. We now show that N fulfills Clause (P3) in the definition of a PI-system. That is, for every d ∈ D N , B N  d is a pure ideal. We show that B N  d fulfills Clauses (S1)–(S3) of Proposition 2.2. / B N . So B N  d is non-principal. So (S1) of Proposition 2.2 holds. By Fact 11, d ∈ Fact 12: B N  d = I ( B M )  d. Let e ∈ B N  d. By Fact 10, e does not have form (I) of Fact 8. So e has form (II). Then by Fact 7, e ∈ I ( B M ). So B N  d ⊆ I ( B M )  d. Clearly, I ( B M )  d ⊆ B N  d. So I ( B M )  d = B N  d. 2 Since I ( B M )  d is a pure ideal in B M , it follows that for every c ∈ B N  d, rk B N (c ) = rk B M (c ) < rk( B N  d). That is, the last clause in the definition of pureness holds.  ( B N ). Then either a ∈ At  ( B M ), or for some i ∈ ω and b, c ∈ I ( B M ), a = (c G \ b) ∪ c. The proof of this fact Fact 13: Let a ∈ At i is left to the reader. 2  ( B N ). If the first case of Fact 13 happens, then We prove that B N  d fulfills Clause (S2) of Proposition 2.2(a). Let a ∈ At applying Proposition 2.4(a) to M, we conclude that a \ d, a ∩ d ∈ B M . So either a \ d ∼ B N a or a ∩ d ∼ B N a. If the second case of Fact 13 happens, then a \ d ∼ B N a. In both cases we found a ∼ B N a such that either a · ( B N  d) = {0} or a ∈ B N  d. That is, B N  d fulfills Clause (S2). We prove that B N  d fulfills (S3). Let β < rk( B N ). Since rk( B N ) = rk( B M ) + 1, we may assume that β = rk( B M ). For every i ∈ ω , c iG ∩ d ∈ I ( B M ). So rk B N (c iG ∩ d) < rk( B M ) = rk B N (c iG ). It follows that rk B N (c iG \ d) = β . Let S = {c iG \ d | i ∈ ω}. Then S is a pairwise disjoint set, and for every s ∈ S, ( B N  d) · s = {0}. So B N  d fulfills (S3). We have shown that B N  d is a pure ideal. This concludes the proof that N is a PI-system. We check that M  N. By Fact 1, I ( B M ) ⊆ I rk( M ) ( B N ). Let a ∈ I ( B N ). By Fact 13, there are ρ , b, c such that ρ ⊆ ω is finite,  b, c ∈ I ( B M ) and a = (( i ∈ρ c iG ) \ b) ∪ c. Suppose that rk N (a) < rk( B M ). Then by Fact 6, ρ = ∅. So a ∈ I ( B M ). We have shown that I rk( M ) ( B N ) ⊆ I ( B M ). Hence I rk( M ) ( B N ) = I ( B M ). So

B M = I rk( M ) ( B M ) ∪ − I rk( M ) ( B M ) = I rk( M ) ( B N ) ∪ − I rk( M ) ( B N ). That is, B M  B N . By definition, D N = D M . Let d ∈ D N . By Fact 12, B N  d = I ( B M )  d. So B N  d = B M  d. We have shown that M  N. Lastly, we show that for every i ∈ ω , Cnvnt( Q i , M i ; N ) holds. Let a ∈ I ( B N ). There is t ∈ N such that a = t G . Let p ∈ G ∩ T t , Q i . So there is c , e  ∈ Q i such that Intrvl(t , p ) ⊆ [c , −e ]. So a = t G ∈ Intrvl(t , p ) ⊆ [c , −e ]. This shows that Cnvnt( Q i , M i ; N ) holds. (b) Let N be as in Part (b). Let G be a directed subset of P N which intersects every dense set of the forms (i) T bN , b ∈ I ( B ); (ii) T iN,k,β , i , k ∈ ω , β < rk( N ); and (iii) T iN,d , i ∈ ω , d ∈ D N . Let d := c 0G .

τ

We show that N [d] is a PI-system. The verification of (P1) and (P2) is trivial. We show that B N  c 0G is a pure ideal. In Part (a) Fact 3 we showed that rk( B N  c 0G ) = rk( B N ). We also showed that B N  c 0G ⊆ I ( B N ). This implies that for every

a ∈ B N  c 0G , rk B N (a) < rk( B N  c 0G ). In particular, B N  c 0G is non-principal. So (S1) of Proposition 2.2 holds.  ( B N ). The fact that G intersects T a implies that a ∩ c G ∈ B N . So either a ∩ c G ∼ B N a and a ∩ c G ⊆ c G or a \ c G ∼ B N a Let a ∈ At 0 0 0 0 0 G and (a \ c 0 ) ∩ c 0G = ∅. This means that (S2) of Proposition 2.2 holds.  β ( B N ) = ∅. Since the c G ’s are pairwise disjoint, it follows that B N  c G Let β < rk( B N ). Then for every i ∈ ω , ( B N  c iG ) ∩ At 0 i fulfills (S3) of Proposition 2.2. Hence B N  c 0G is a secluded ideal. By the above paragraph, B N  c 0G is also pure. We have shown that N [d] fulfills (P3) in the definition of a PI-system. Let e ∈ D N . Choose p ∈ G ∩ T 0,e and let 0, e , b ∈ σ p . Then c 0G ∩ e = b ∈ B N . Hence (P4) is fulfilled. This proves (b). / B N . This has been already (c) Part (c) holds almost by definition. The only fact that needs verification is that c 0G ∈ checked. 2 Proof of Theorem 2.7 (Construction Theorem). For a well-ordered set L, Lim( L ) denotes the set of limit points of L. So Lim(ω1 ) = {ω · α | 0 = α ∈ ω1 }. We define by induction on α ∈ Lim(ω1 ) the following objects: (1) A countable PI-system M α such that B ∗M α is a subalgebra of P (ω) and At( B ∗M α ) = {{n} | n ∈ ω}. (2) A dense set of intervals Q α of M α . (3) A bijection f α : α → B M α .

R. Bonnet, M. Rubin / Topology and its Applications 158 (2011) 1503–1525

Let { S α | α < ω1 } be a ♦-sequence for ω1 . The induction hypotheses are as follows. Suppose that Lim(γ ). Then for every α , β ∈ Lim(γ ) such that α  β : (H1) (H2) (H3) (H4)

1513

γ ∈ Lim(ω1 ) and M α , Q α , f α have been defined for every α ∈

If α = ω · α  , then rk( M α ) = α  . Mα  Mβ . Cnvnt( Q α , M α ; M β ) holds. fα ⊆ fβ .

The case γ = ω : Take D ω to be any infinite partition of ω into infinite sets. Let B ω be the subalgebra of P (ω) generated by {{n} | n ∈ ω} ∪ D ω , B ω be the algebra of finite and cofinite subsets of ω and M ω =  B ω , B ω , D ω . Set Q ω = Wip( M ω ) and let f ω be a bijection between ω and B ω . It is obvious that the induction hypotheses hold. The limit  case: Let γ ∈ Lim(Lim (ω1 )), and suppose that for every M γ = α ∈Lim(γ ) M α and f γ = α ∈Lim(γ ) f α .

α ∈ Lim(γ ), M α , Q α and f α have been defined. Let Pγ

Let P γ := f γ ( S γ ). If P γ ⊆ I ( B M γ ), and is nowhere dense in M γ , let Q γ := Q M γ , that is,





Q γ = c , e  ∈ Wip( M γ ) P γ ∩ [c , −e ] = ∅ . Otherwise, let Q γ := Wip( M γ ). It follows from the Limit Case Proposition 2.13 that M α , Q α , f α ,

α ∈ Lim(γ + ω) satisfy the induction hypotheses.

The successor case: Let γ be a successor in Lim(ω1 ), and suppose that M β , Q β and f β have been defined for every β ∈ Lim(γ ). Let α ∈ Lim(ω1 ) be such that γ = α + ω . By Part (a) of the Successor Case Lemma 2.12, there is a countable M γ  M α such that rk( M γ ) = rk( M α ) + 1, D M γ = D M α , and for every limit β ∈ Lim(γ ), Cnvnt( Q β , M β ; M γ ) holds. Let dγ ⊆ ω and M γ [dγ ] be as assured by Part (b) of the Successor Case Lemma 2.12 applied to M γ , and let M γ :=

M γ [dγ ].

We now define f γ and Q γ . Let f γ be a bijection between I ( B M γ ), and is nowhere dense in M γ , let Q γ :=

Pγ Q Mγ

γ and B M γ such that f γ ⊇ f α . Let P γ := f γ ( S γ ). If P γ ⊆

. Otherwise, let Q γ := Wip( M γ ).

It follows from the three parts of the Successor Case Lemma 2.12 and from Proposition 2.10(d) that the induction hypotheses hold.  Let M = α <ω1 M α . We show that M satisfies (C1)–(C4) of Theorem 2.7. (C1) By Proposition 2.10(e), for every α ∈ Lim(ω1 ), M α  M. Since for every α = ω · α  ∈ Lim(ω1 ), M α is countable and rk( M α ) = α  , we conclude that rk( M ) = ω1 and B M is a thin–tall BA. Hence (C1) holds. (C2) Let d ∈ D M . Hence for some α ∈ Lim(ω1 ), d ∈ D M α . Since M α  M, B M  d = B M α  d. Hence rk( B M  d) = rk( B M α  d)  α . So (C2) holds. (C3) Let α ∈ Lim(ω1 ) be a successor element of Lim(ω1 ). By our construction, dα is defined and rk( B M  dα ) = rk( M α  dα ) = α  . So (C3) holds. (C4) holds. Suppose by contradiction that P ⊆ I ( B M ) is uncountable and nowhere dense. Let N :=  B M , B M , D M , I ( B M ), I M , P . That is, N is the expansion of B M , obtained by adding unary predicates which represent B M , D M , I ( B M ), I M and P (so the universe of N is B M , and for instance, N | P (x) if and only if x ∈ P ). For α ∈ Lim(ω1 ) let N α be the  substructure of N whose universe is B M α . Then C := {α ∈ Lim(ω1 ) | N α ≺ N } is a closed unbounded subset of ω1 . Let f = { f α | α ∈ Lim(ω1 )}.

By ♦ℵ1 , S := {α ∈ ω1 | f −1 ( P ) ∩ α = S α } is stationary in ω1 . Denote the universe of any structure N  by | N  |. Let β ∈ S ∩ C . The fact that P is nowhere dense in M is expressible by a first order sentence in the language of N. Since N β ≺ N, the same sentence holds in N β , so P ∩ | N β | is nowhere dense in M β . Since β ∈ S, f −1 ( P ) ∩ β = S β . So f β ( S β ) = f ( S β ) = P ∩ f (β) = P ∩ | N β |. That is, f β ( S β ) is nowhere dense Pβ

in M β . So P β = f ( S β ) and Q β = Q M β . Let a ∈ P \ | N β |. Since a ∈ I ( B M ) \ B M β and Cnvnt( Q β , M β ; M ) holds, there is c , e  ∈ Q β such that a ∈ [c , −e ]. By the definition of Q β , P β ∩ [c , −e ] = ∅. So N β | ¬∃x( P (x) ∧ (x ∈ [c , −e ])), whereas N | ∃x( P (x) ∧ (x ∈ [c , −e ])). (This is so, since P (a) ∧ (a ∈ [c , −e ]) holds.) This contradicts the fact that N β ≺ N. So M satisfies (C4). 2 We conclude this section by observing that there are 2ℵ1 pairwise non-isomorphic condensed BA’s. Let B, C be uncountable Boolean algebras. We say that B and C are far, if there is no uncountable BA which is embeddable in both B and C . Note that since condensed BA’s are downward-categorical, being non-isomorphic implies being far. We may also define the notion of quotient-far. B and C are quotient-far, if they do not have isomorphic uncountable quotients. We prove a little more. Namely, there is a family B of condensed Boolean algebras such that |B | = 2ℵ1 and for every distinct B , C ∈ B , B and C are far and quotient-far.

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Let B  C mean that there are countable ideals I , J of B and C such that B / I ∼ = C / J . Note that  is an equivalence relation on the class of Boolean algebras and that for sub-Ostaszewski Boolean algebras, B  C iff there are ideals I , J of B and C such that B / I is uncountable and B / I ∼ = C/ J. Theorem 2.17. (♦ℵ1 ) There is a set B of condensed BA’s such that |B | = 2ℵ1 , and for every distinct B 1 , B 2 ∈ B , B 1  B 2 . Proof. Let T = {τ | there is α ∈ ω1 such that τ : α → {0, 1}}, T (<β) = {τ ∈ T | Dom(τ ) < β}, T (β) = {τ ∈ T | Dom(τ ) = β} and  T = {0, 1}ω1 . We shall construct a family of countable PI-systems { M τ | τ ∈ T }. The construction takes care that for every  σ ∈ T , M σ := α ∈ω1 M σ α is well defined and is a condensed PI-system. For σ ∈ T ∪  T denote B M σ by B σ . Then B σ is made to be a subalgebra of P (ω) and At( B σ ) = {{n} | n ∈ ω}. Assume temporarily that the construction has the property that for every distinct σ , ν ∈  T , B σ = B ν . Suppose that f : Bσ ∼ T, = B ν . Then there is π f ∈ Sym(ω) such that f (a) = π f [a] for every a ∈ B σ . This implies that for every σ ∈  |{ν ∈  T | Bν ∼ T }, such that |B | = 2ℵ1 , and for every = B σ }|  2ℵ0 = ℵ1 . It follows that there is a subset B ⊆ { M σ | σ ∈  distinct B 1 , B 2 ∈ B , B 1  B 2 . T , B σ  B ν . This can be done assuming CH, but We have to do more in order to obtain that for every distinct σ , ν ∈  since we have to use ♦ elsewhere in the proof, we also use ♦ here. We now construct { M σ | σ ∈ T }. In fact, for every σ ∈ T we shall have the following objects: (1) A countable PI-system M σ . Set B σ := B M σ and B ∗σ := B ∗M σ . Then B σ is a subalgebra of P (ω) and At( B σ ) = {{n} | n ∈ ω}. (2) A dense set of intervals Q σ of M σ . (3) A bijection f σ : ω · α → B ∗σ , where α = Dom(σ ). Let Let

α ∈ ω1 and suppose that M σ has been constructed for every σ ∈ T (<α ) . We have the following induction hypotheses. τ ∈ T (α ) . So the sequence { M τ β | β < α } is defined. We assume that the sequence ( M τ β , Q τ β , f τ β ),

β <α

fulfills Clauses (H1)–(H4) from the proof of Theorem 2.7. More precisely, denote M τ β , Q τ β and f τ β by respectively M ω·β , Q ω·β and f ω·β . Then  ω·β ω·β ω·β  M ,Q ,f , β <α fulfills Clauses (H1)–(H4) from the proof of Theorem 2.7. Let { S α | α ∈ ω1 } be a ♦-sequence. We may assume that for every

α ∈ ω1 , S α is an object of the following form:

 Q α , ηα , ζα , hˆ α , Iα , J α , α ,  where Q I α and  J α , are subsets of α , hˆ α is a subset of α × α , and ηα and ζα are subsets of α × {0, 1}. (At stage α of the construction we shall use S ω·α .) Recall that at stage α of the construction we have already constructed { M σ | σ ∈ T (<α ) }, and we construct { M σ | σ ∈ T (α ) }. Case 1: α = 0. Define M Λ as M ω was defined in the proof of Theorem 2.7. Case 2: α ∈ Lim(ω1 ). Let τ ∈ T (α ) . Set γ = ω · α . Recall that the sequence ( M τ β , Q τ β , f τ β ), β < α , fulfills Clauses (H1)–(H4) from the proof of Theorem 2.7. In this case we define M τ , f τ as in the limit case in the proof of γ . That is, define P τ := f τ ( Q γ ). If P τ ⊆ I ( B M τ ), and is nowhere Theorem 2.7, and we define Q τ as in Theorem 2.7 using Q dense in M τ , then let





Q τ = Q Mττ := c , e  ∈ Wip( M τ ) P τ ∩ [c , −e ] = ∅ . P

Otherwise, let Q τ := Wip( M τ ). Case 3: For some δ ∈ Lim(ω1 ), α = δ + 1. Set γ = ω · α . We distinguish between two subcases. γ , ηγ , ζγ , hˆ γ ,  , η, ζ, hˆ , Case 3.1: Recall that B σ := B M σ and set B ∗σ := B ∗M σ . Denote  Q Iγ , J γ  by  Q I, J . Assume that: (1) η, ζ ∈ {0, 1}δ , η = ζ . I ) is an ideal in B η and for some β < δ , I ⊆ I β ( B η ). (2) I := f η ( (3) J := f ζ ( J ) is an ideal in B ζ and for some β < δ , J ⊆ I β ( B ζ ). (4) Set h = f ζ ◦ hˆ ◦ f η−1 . Then h : B η / I ∼ = Bζ / J . Case 3.2: Case 3.1 does not hold. We start with the second case. Case 3.2. Let τ ∈ {0, 1}α . Define M τ , Q τ and f τ as in the successor case of Theorem 2.7. We give more details. For every β  δ define M ω·β = M τ β , Q ω·β = Q τ β and f ω·β = f τ β . Then the sequence

 ω·β ω·β ω·β  M ,Q ,f ,

β δ

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satisfies Clauses (H1)–(H4) from the proof of Theorem 2.7. Let M ω·α and f ω·α be as assured by the successor case of ω·α ). If P τ ⊆ I ( B M τ ), and is nowhere dense in M τ , Theorem 2.7. We then set M τ = M ω·α and f τ = f ω·α . Define P τ := f τ ( Q then let





Q τ = Q Mττ := c , e  ∈ Wip( M τ ) P τ ∩ [c , −e ] = ∅ . P

Otherwise, let Q τ := Wip( M τ ). Case 3.1. Let τ ∈ {0, 1}α . If τ  δ ∈ / {η, ζ }, then define M τ , Q τ and f τ as in Case 3.2. It remains to define M η ˆ  , M ζ ˆ  , etc. for = 0, 1. In fact, we shall have that M η ˆ0 = M η ˆ1 and M ζ ˆ0 = M ζ ˆ1 . The sequence {( M ηβ , Q ηβ , f ηβ ) | β  δ} satisfies Clauses (H1)–(H4) from the proof of Theorem 2.7. Let ( M η , Q η , f η ) be as in the successor case of Theorem 2.7. For ζ too, define ( M ζ , Q ζ , f ζ ) as in Theorem 2.7. Define

M ηˆ0 = M ηˆ1 = M η ,

Q ηˆ0 = Q ηˆ1 = Q η

and

f ηˆ0 = f ηˆ1 = f η .

Set B η = B M η and B ζ = B M  . ζ Suppose first that there is no g : B η / I ∼ = B ζ / J such that g ⊇ h. Then define

M ζ ˆ0 = M ζ ˆ1 = M ζ ,

Q ζ ˆ0 = Q ζ ˆ1 = Q ζ

and

f ζ ˆ0 = f ζ ˆ1 = f ζ .

 δ ( B  ) be such that Now assume that there is g : B η / I ∼ = B ζ / J such that g ⊇ h. Write M ζ =  B ζ , B ζ , D ζ . Let A ⊆ At ζ B

B ζ = cl( B ζ ∪ A ), and for every distinct a, b ∈ A, a ∼ ζ b. Let a0 ∈ A. Define B = cl( B ζ ∪ ( A \ {a0 })) and B ∗ = cl( B ∪ D ζ ). Let N =  B ∗ , B , D ζ . Then M ζ  N. Since B ∗ ⊆ B ∗M  , it follows that for every β  δ , Cnvnt( Q ζ β , M ζ β ; N ) holds. We use the ζ

Successor Case Lemma 2.12. Let d ∈ P (ω) be as assured by Lemma 2.12(b). Then by Lemma 2.12(b) and (c), M ζ  N [d], and for every β  δ , Cnvnt( Q ζ β , M ζ β ; N [d]) holds. Define

M ζ ˆ0 = M ζ ˆ1 = N [d]. We then define f ζ ˆ0 , Q ζ ˆ0 , f ζ ˆ1 and Q ζ ˆ1 as in Case 3.2. Case 4: α = 1 or for some β ∈ ω1 , α = β + 2. For τ ∈ {0, 1}α define M τ , Q  τ and f τ as in Case 3.2. This concludes the definition of { M τ | τ ∈ T }. Recall that for σ ∈  T , M σ = α <ω1 M σ α and B σ = B M σ . For every

σ ∈ T,  M σ is condensed. This is so, since for every σ ∈ T , the sequence { M σ α | α < ω1 } fulfills Clauses (H1)–(H4) from the proof

of Theorem 2.7. We show that for every distinct

σ,τ ∈ T , B σ  B τ .

Claim 1. Suppose that M ηˆ j  and M ζ ˆ  were obtained from the construction of Case 3.1. That is: (1) η, ζ ∈ {0, 1}δ , η = ζ . I ) is an ideal in B η and for some β < δ , I ⊆ I β ( B η ). (2) I := f η ( (3) J := f ζ ( J ) is an ideal in B ζ and for some β < δ , J ⊆ I β ( B ζ ). (4) Set h = f ζ ◦ hˆ ◦ f η−1 . Then h : B η / I ∼ = Bζ / J .

Set ηˆ = η ˆ  j  and ζˆ = ζ ˆ  . Then there is no f : B ηˆ / I ∼ = B ζˆ / J such that f ⊇ h. Proof. Let B η and B ζ be as in Case 3.1. If there is no g : B η / I ∼ = B ζ / J such that g ⊇ h, then Claim 1 is true since B ηˆ = B η and B ζˆ = B ζ . Now assume that g : B η / I ∼ = B ζ / J and g ⊇ h. Assume by contradiction that there is f : B ηˆ / I Let A, a0 and B be as in Case 3.1. Then B ζˆ = B ⊆ B ζ . Hence k := f ◦ g −1 is an embedding of

∼ = B ζˆ / J such that f ⊇ h. B ζ / J in itself. Also, since f , g extend h and Dom(h) = B ζ / J , it follows that k  ( B ζ / J ) = Id. Since for some β < δ , J ⊆ I β ( B ζ ), At( B ζ / J ) ⊆ B ζ / J . Hence

k  At( B ζ / J ) = Id. So k = Id. We show that k(a0 / J ) = a0 / J . Clearly, k(a0 / J ) ∈ B / J , for every b ∈ a0 / J , b ∼ no b ∈ B such that b ∼

B ζ

a0 . So k(a0 / J ) = a0 / J . A contradiction.

B ζ

a0 and there is

2

 We prove the theorem.  For every τ ∈ T the sequence { M τ α | α < ω1 } fulfills Clauses (H1)–(H4) from Theorem 2.7. Also, M τ = { M τ α | α < ω1 }. So M τ is condensed. Suppose by contradiction that there are distinct σ , τ ∈  T , countable ideals I ⊆ B σ , J ⊆ B τ , and h such I∼ I = f σ−1 [ I ],  J = f τ−1 [ J ] and hˆ = = B τ / J . Let γ ∈ ω1 be such that rk( I ), rk( J ) < γ and σ  γ = τ  γ . Let  Recall that Sα =  Q α , ηα , ζα , hˆ α , Iα , J α .

the proof of that h : B σ / f τ−1 ◦ h ◦ f σ .

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There is δ ∈ ω1 such that γ + δ = δ , δ = ω · δ , ηδ = σ  δ , ζδ = τ  δ ,  I δ = I ∩ δ,  Jδ =  J ∩ δ and hˆ δ = hˆ  δ . Then M σ (δ+1) and M τ (δ+1) are constructed according to Case 3.1. By Claim 1,

(∗) there is no f such that f : B σ (δ+1) / I ∼ = B τ (δ+1) / J and f ⊇ h  ( B σ δ / I ). Note that I δ+1 ( B σ / I ) = I δ+1 ( B σ )/ I . This follows from the fact that rk( I ) + δ = δ . The same holds for I δ+1 ( B σ (δ+1) ) and the same holds for τ . So we have



τ . Also, I δ+1 ( B σ ) =



h I δ+1 ( B σ / I ) = I δ+1 ( B τ / J ) and

I δ+1 ( B σ / I ) = I δ+1 ( B σ )/ I = I δ+1 ( B σ (δ+1) )/ I . Hence





h I δ+1 ( B σ (δ+1) )/ I = I δ+1 ( B τ (δ+1) )/ J . Since B σ (δ+1) is generated by I δ+1 ( B σ (δ+1) ) and the same holds for

τ,

h[ B σ (δ+1) / I ] = B τ (δ+1) / J . This means that

∼ B τ (δ+1) / J . h  ( B σ (δ+1) / I ) : B σ (δ+1) / I = But h  ( B σ (δ+1) / I ) ⊇ h  ( B σ δ / I ). This contradicts (∗).

2

3. Downward-categoricity of condensed Boolean algebras The main goal of this section is the following theorem. Theorem 3.1. If B is a condensed Boolean algebra, then B is downward-categorical. Additional properties of condensed Boolean algebras are proved in Section 4. Definition 3.2. (a) Let B be a Boolean algebra and I be a proper ideal in B. Then B± ( I ) denotes I ∪ − I . Recall that B± ( I ) B does not depend on B. Set C± ( I ) := B± (cmpl B ( I )). (b) Let B be a Boolean algebra. B is called a rich Boolean algebra, if it is thin–tall and unitary, and for every α < ω1 there is a pure ideal I of B such that α  rk( I ) < ω1 . The fact that a condensed BA is downward-categorical follows from the following two lemmas. Lemma 3.3. Let B be a rich Boolean algebra and J be a secluded ideal in B. Assume that rk( J ) < ω1 . Let C be a subalgebra of B such B ( J ). Then C ∼ that C ⊇ C± = B. Lemma 3.4. Let B be a condensed Boolean algebra, and C be an uncountable subalgebra of B. Then there is a pure ideal J in B such B ( J ). that rk( J ) < ω1 and C ⊇ C± Proof of Theorem 3.1 assuming Lemmas 3.3 and 3.4. The fact that B is a thin–tall algebra is trivial. Let C be an uncountable B subalgebra of B. By Lemma 3.4, there is an ideal J of B such that J is pure, rk( J ) < ω1 and C ⊇ C± ( J ). By the definitions of a condensed PI-system and of a PI-system (see (C2), (C3) and (P3)), B is rich. So by Lemma 3.3, C is isomorphic to B. 2 Definition 3.5. (a) For a countable ordinal α let Bα denote the unique (up to isomorphism) unitary countable superatomic BA with rank α . (b) For unitary BA’s B 1 and B 2 we denote by B 1  B 2 the subalgebra of B 1 × B 2 generated by I ( B 1 ) × {0 B 2 } ∪ {0 B 1 }× I ( B 2 ).

Lemma 3.6. (a) Let B be a thin–tall Boolean algebra. B (1) Let J be a pure ideal in B. Then B± ( J ) is a unitary Boolean algebra, C± ( J ) is a thin–tall Boolean algebra, and B ∼ = B C± ( J )  B± ( J ).

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(2) If J ⊆ B is a pure ideal with rank β , then B ± ( J ) ∼ = Bβ . B (3) Suppose that B is rich, and J is a secluded ideal in B such that rk( J ) < ω1 . Then C± ( J) ∼ = B. (b) Let B be a rich BA, C be a subalgebra of B, and E be an ideal in B such that either E is principal generated by a member of I ( B ), or E is a secluded ideal and rk( E ) < ω1 . Suppose that C ⊇ cmpl B ( E ). Then C is rich. Proof. (a1) It follows trivially from the definition of a pure ideal that B± ( J ) is unitary and that I (B± ( J )) = J . It is also B B ( J ) is a thin–tall BA, and that I (C± ( J )) = cmpl B ( J ). So trivial that C±









B C± ( J )  B± ( J ) = a, b a ∈ cmpl B ( J ) and b ∈ J ∪ −a, −b a ∈ cmpl B ( J ) and b ∈ J . B Let f : C± ( J )  B± ( J ) → B be defined as follows:





f a, b =



a+b

if a ∈ cmpl B ( J ) and b ∈ J ,

a·b

if a ∈ −cmpl B ( J ) and b ∈ − J .

B It left to the reader to show that f is an isomorphism between C± ( J )  B± ( J ) and B. (a2) The proof of (a2) is trivial. B ( J ) is a thin–tall BA. Let I be a pure ideal of B such that rk( J ) < rk( I ) < ω1 . Set β = rk( I ). (a3) It is trivial that C := C±

Let I 1 := I ∩ cmpl B ( J ). It is easy to see that I 1 is a pure ideal in both B and C , and that rk( I 1 ) = β . Let J 1 be the ideal generated by J ∪ I 1 . By Proposition 2.2(e), J 1 is a pure ideal and rk( J 1 ) = β . So by (a1) and (a2), C C C∼ ( I 1 )  B± ( I 1 ) ∼ ( I 1 )  Bβ = C± = C±

and

B B B∼ ( J 1 )  B± ( J 1 ) ∼ ( J 1 )  Bβ . = C± = C±

C B A trivial computation shows that C± ( I 1 ) = C± ( J 1 ). Hence that B ∼ = C . We make this computation. We just have to show that

cmplC ( I 1 ) = cmpl B ( J 1 ). Indeed, a ∈ cmplC ( I 1 ) ⇔ a ∈ C and a · I 1 = {0} ⇔ (i) a · J = {0} and a · I 1 = {0} or (ii) −a · J = {0} and a · I 1 = {0}. However, since both I 1 and J are secluded ideals in B, there is no a ∈ B satisfying (ii). So the above is equivalent to: (i) a · J = {0} and a · I 1 = {0}. Clearly, (i) ⇔ a ∈ cmpl B ( I ∪ J ) ⇔ a ∈ cmpl B ( I 1 ). (b) Let I be a pure ideal in B such that rk( I ) > rk( E ). Then I ∩ cmpl B ( E ) is a pure ideal in C and rk( I ∩ cmpl B ( E )) = rk( I ). This implies that C is rich. 2 B Proof of Lemma 3.3. Let B, J and C be as in Lemma 3.3. Let C 1 := C± ( J ). By Lemma 3.6(a3), C 1 ∼ = B. We show that C 1 ∼ = C.

The fact C ⊇ cmpl B ( J ) and Lemma 3.6(b), imply that C is a rich. Since C 1 ∼ = B, C 1 is rich. Let J 1 := J ∩ C . Choose J 2 ⊆ C 1 such that J 2 is a pure ideal in C 1 and rk( J 2 ) > rk( J 1 ). Fact 1: J 2 is a pure ideal in C . We have that J 2 ⊆ I (C 1 ) and that I (C 1 ) is an ideal in C . So J 2 is an ideal in C . We now check properties (S1)–(S3) of Proposition 2.2(a). Obviously J 2 is non-principal and for every β < ω1 , there is b ∈ cmplC ( J 2 ) such that rk(b) = β . So (S1) and (S3) hold. Let c ∈ I (C ). Write c as c = c 1 + c 2 , where c 1 ∈ J and c 2 ∈ cmplC ( J ). Then c 2 ∈ C 1 . Moreover, c 2 ∈ I (C 1 ). So there are c 3 ∈ cmplC ( J 2 ) and c 4 ∈ J 2 such that c 2 = c 3 + c 4 . Then c = c 1 + c 3 + c 4 , c 1 + c 3 ∈ cmplC ( J 2 ) and c 4 ∈ J 2 . So (S2) holds. Also, for every a ∈ J 2 , rkC (a) = rkC 1 (a) < rk( J ). Hence J 2 is a pure ideal in C . 2 Fact 2: Either J 1 is a principal ideal in C , or J 1 is a secluded ideal in C . Suppose that J 1 is not principal. Then (S1) holds. Clearly, rk( J 1 )  rk( J ) < ω1 = rk(C ). So (S3) holds. Let c ∈ I (C ). Then c ∈ I ( B ). Since J is either secluded or principal, there are c 1 ∈ J and c 2 ∈ cmpl B ( J ) such that c = c 1 + c 2 . Since C ⊇ cmpl B ( J ), it follows that c 2 ∈ C . Hence c 1 ∈ C . So c 1 ∈ J 1 and c 2 ∈ cmplC ( J 1 ). That is, (S2) holds for J 1 and C . Hence J 1 is a secluded ideal in C . 2 Let J 3 be the ideal of C generated by J 1 ∪ J 2 . Then by Proposition 2.2(e), J 3 is a pure ideal in C . By Lemma 3.6(a3), C C± ( J 3) ∼ =C

and

C±1 ( J 2 ) ∼ = C1. C

C A trivial computation shows that cmplC ( J 3 ) = cmplC 1 ( J 2 ). So C± ( J 3 ) = CC 1 ( J 2 ). It follows that C ∼ = C 1 . Since C 1 ∼ = B, it follows that C ∼ = B. 2

To prove Lemma 3.4, we need another fact which will be used frequently. Proposition 3.7.

/ I ( B M ), then [c , −e ] B M ⊆ I ( B M ). Also, every wide interval (a) Let M be any PI-system and [c , −e ] be a wide interval of M. If e ∈ BM contains a wide interval [c , −e ] such that e ∈ / I ( B M ). (b) Let B be a condensed Boolean algebra and M be a PI-system such that B = B M . Any meet-closed uncountable subset P of I ( B ) contains a wide interval.

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Proof. (a) The trivial proof is left to the reader. / I ( B ). Let (b) Let [c 0 , −e ] B be a wide interval such that P is dense in [c 0 , −e ] B . By Part (a), we may assume that e ∈ c ∈ P ∩ [c 0 , −e ] B . Then P is dense in [c , −e ] B . We show that [c , −e ] B ⊆ P . Let d ∈ [c , −e ] B . By Part (a), d ∈ I ( B ). So [d, −e ] B is a wide interval. Choose a ∈ P ∩ (d, −e ] B and let e  = e + (a − d). Then d, e   ∈ Wip( M ). Choose b ∈ [d, −e  ] B ∩ P . We have a · b = d and a, b ∈ P . Hence d ∈ P . That is [c , −e ] B ⊆ P . 2 Proof of Lemma 3.4. Let B be a condensed BA and M be a condensed PI-system such that B = B M . Suppose that C is an uncountable subalgebra of B. So C ∩ I ( B ) is uncountable and meet-closed. By Proposition 3.7(b), there is c , e  ∈ Wip( M ) such that [c , −e ] ⊆ C ∩ I ( B ). Let e 0 = c + e and J 1 = B  e 0 . We show that C ⊇ cmpl B ( J 1 ). Since c ∈ [c , −e ], we have that c ∈ C . Let b ∈ cmpl B ( J 1 ). Hence b  −e. So c + b ∈ [c , −e ] ⊆ C . Also, b · c = 0. We conclude that b = (c + b) − c ∈ C . That is, C ⊇ cmpl B ( J 1 ).  By Proposition 2.10(a) e 0 has the form e 0 = ( D 0 − u ) + v, where D 0 is a finite subset of D M and u , v ∈ I ( B ). By the definition of a condensed PI-system, rk( B  d) < ω1 for every d ∈ D 0 . So rk( J 1 ) < ω1 . It also follows from Proposition 2.10(a) that J 1 is secluded or that J 1 is principal. Let d ∈ D M be such that rk( B  d) > rk( J 1 ). Let J 2 = B  d and J be the ideal generated by J 1 ∪ J 2 . Then by Lemma 3.6(a3), J is pure and rk( J ) < ω1 . Clearly, cmpl B ( J ) ⊆ cmpl B ( J 1 ). So C ⊇ cmpl B ( J ). 2 We shall later see that a condensed Boolean algebra is never quotient-categorical. It is true however that, assuming CH, every uncountable quotient of a condensed BA is condensed. We need the following fact. Corollary 3.8. Let B be a condensed BA and I ⊆ B be an uncountable ideal. Then | B / I |  ℵ0 . Proof. If I ⊆ I ( B ), then it is trivial that | B / I |  ℵ0 . So suppose that I ⊆ I ( B ). By Lemma 3.4, there is a countable pure ideal J B ( J ). Clearly, cmpl B ( J ) ⊆ I ( B ) and (− I ) ∩ I ( B ) = ∅. So I ⊇ cmpl B ( J ). It follows from the pureness of J such that B± ( I ) ⊇ C± that for every a ∈ I ( B ) there is a ∈ J such that a / I = a/ I . So | B / I |  | J | = ℵ0 . 2 Theorem 3.9. (a) Let B be a narrow condensed BA, and I be an ideal of B such that B / I is uncountable. Then B / I is condensed. (See Remark 2.8.) (b) (CH) Let B be a condensed BA and I be an ideal of B such that B / I is uncountable. Then B / I is condensed. Proof. We prove (a) and (b) together. Let M =  B  , B , D  be a condensed PI-system. By Corollary 3.8, | I |  ℵ0 , and hence I ⊆ I ( B ). Let β = rk( I ) · ω . Set T := {d ∈ D | rk( B  d)  β} and S := D \ T . Then | S |  2ℵ0 = ℵ1 . Let π : S → T be 1–1. For every s ∈ S let e s = π (s) + s and set E = {e s | s ∈ S } ∪ ( T \ Rng(π )). Since I ⊆ I ( B ), it follows that I is an ideal of B  . For every b ∈ B  , let b I denote b/ I , and for every subset A ⊆ B  , denote A I := {a I | a ∈ A }. Let  B be the subalgebra of B  generated by I I I  B ∪ E. Define N =  B , B , E . We show that N is a condensed PI-system. Note first that for every e ∈ E, (A1) rk( B I  e I ) = rk( B  e ) and (A2) B  e is a pure ideal in B. B I is generated by B I ∪ E I . By (A1), E I is We start with properties (P1)–(P4) of a PI-system. Clearly, B I is unitary and  infinite. We show that B I is dense in  B I . (This implies that  B I is atomic and that At( B I ) = At( B I ).)  Let A = {e − a | e ∈ E , a ∈ I ( B )}. For every y ∈ B \ B: either (i) y is the finite sum of elements from A or (ii) − y is the finite sum of elements from A. Suppose that (i) happens, and let e − a  y. Then there is b ∈ I ( B )  (e − a) such that rk B (b) > β . So 0 = b I  y I and b I ∈ B I . If (ii) happens, then A  y = ∅. As in Case (i), it follows that B I  y I = ∅. We have shown that B I is dense in  BI. I I / B I . We have shown Let e ∈ E. By (A1), for every a ∈ B  e I , rk B (a I ) < rk( B I  e I ). So B I  e I is non-principal. Hence e I ∈ that (P1) and (P2) hold. Now, for every e ∈ E, B I  e I is a pure ideal of B I . This follows from the non-principality of B I  e I and from (A2). So (P3) holds. For every distinct e , f ∈ E, e · f ∈ B. So the same is true for e I , f I and B I . So (P4) holds. We have shown that N is a PI-system, and it remains to show that N fulfills properties (C1)–(C4) of a condensed PI-system. Properties (C1)–(C3) are trivial, and we prove (C4). Let A ⊆ B be uncountable. We show that A I is somewhere . There is a wide interval L := [c , − f ] dense. Define ρ : D → E as follows: ρ (d) = d + π (d) if d ∈ S, and ρ (d) = d if d ∈ T of M such that A is dense in L. And L has a wide subinterval of the form [c 0 , −( D 0 − b)] , where c 0 ∈ I ( B ) and D 0 is a finite subset of D. Then K := [c 0 , −( d∈ D 0 ρ (d) − b − c 0 )] is a wide subinterval of [c 0 , −( D 0 − b)], and hence A is



dense in K . The set [c 0I , −(

d∈ D 0

ρ (d) I − b I − c0I )] is a wide interval of N and A I is dense in this interval. 2

4. Rigidity and some other properties A condensed BA is as rigid as a downward-categorical thin–tall BA could be. We explain this statement. A downwardcategorical thin–tall BA must have countable secluded ideals. To see this, let B be a downward-categorical thin–tall BA. Let

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1519

 α ( B ). Let A be the subalgebra of B generated by I α ( B  a) ∪ I ( B  −a). Since A is uncountable, B ∼ α < ω1 and a ∈ At = A. Also, I α ( B  a) is a pure ideal in A of rank α . This implies that B has pure ideals of unbounded rank. Let B be a thin–tall BA and f be an endomorphism of B. We say that f is a trivial endomorphism if there is a countable secluded ideal I of B such that f (a) = a for every a ∈ cmpl B ( I ). We consider a more general situation. Let B be a thin–tall BA, I ⊆ B be an ideal and f : B → B / I be a homomorphism. We say that f is a trivial homomorphism if there is a countable secluded ideal J of B such that f (a) = a/ I for every a ∈ cmpl B ( J ). In this section we prove that if B is a condensed Boolean algebra, and f is an endomorphism of B such that |Rng( f )| = ℵ1 , then f is a trivial endomorphism. In fact, we prove the analogous claim for homomorphisms. Lemma 4.1. Let B a condensed BA, I ⊆ B be an ideal and f : B → B / I be a homomorphism such that |Rng( f )| = ℵ1 . Let





A = a ∈ B for every b ∈ f (a), rk(b  a)  rk(a) . Then | A |  ℵ0 . Proof. Let B  and D be such that M =  B  , B , D  is a condensed PI-system. We show that for every a ∈ I ( B ) and b ∈ B, if f (a) = b/ I , then b ∈ I ( B ). Suppose by contradiction that a and b refute the above. So

f [ B  −a] ⊆ ( B / I )  − f (a) = {c / I | c  −b}. Since B  −b is countable, f [ B  −a] is countable. Since B  a is countable, it follows that f [ B  a] is countable. Hence Rng( f ) is countable. A contradiction. Since B / I is uncountable, I is countable. Let α be such that I ⊆ I α ( B ). Set





C = a ∈ I ( B ) for every b ∈ f (a), rk(a − b) = rk(a) ,

 E = a ∈ I ( B ) for every b ∈ f (a), rk(b − a)  rk(a) .

We show that A ⊆ C ∪ E ∪ I α ( B ). Let a ∈ A \ (C ∪ E ). We show that rk(a) < α . There is c ∈ f (a) such that rk(a − c ) < rk(a). There is e ∈ f (a) such that rk(e − a) < rk(a). Since a ∈ A, rk(c − a)  rk(a),

c − e = c − (e ∩ a) − (e − a). Now, c − (e ∩ a) ⊇ c − a. Hence rk(c − (e ∩ a))  rk(c − a)  rk(a). Recall that rk(e − a) < rk(a). So











rk(c − e ) = rk c − (e ∩ a) − (e − a) = rk c − (e ∩ a)  rk(a). Since c − e ∈ I , rk(c − e ) < α . Hence rk(a) < α . That is, a ∈ I α ( B ). We have shown that A ⊆ C ∪ E ∪ I α ( B ). We leave it to the reader to show that |C |, | E |  ℵ0 . Since A ⊆ C ∪ E ∪ I α ( B ), | A |  ℵ0 . 2 Proposition 4.2. Let B a condensed BA, I ⊆ B be an ideal and f : B → B / I be a homomorphism such that |Rng( f )| = ℵ1 . Then  ( B ) | f (a) = a/ I }| = ℵ1 . |{a ∈ At Proof. By Lemma 4.1, there is

α such that:

• for every a ∈ I ( B ) \ I α ( B ) there is ba ∈ f (a) such that rk(a  ba ) < rk(a).  β ( B ). Denote baβ by bβ . By Fodor’s Lemma, there is an uncountable set For every countable β  α choose some aβ ∈ At A ⊆ ω1 and a ∈ B such that for every β ∈ A, aβ  bβ = a. Let γ = min( A ) and for every β ∈ A \ {γ } let c β = aβ  aγ and  β ( B ) and dβ ∈ f (c β ). Now, dβ = bβ  bγ . Then c β ∈ At c β  dβ = (aβ  aγ )  (bβ  bγ ) = (aβ  bβ )  (aγ  bγ ) = a  a = 0. That is, c β = dβ . So c β ∈ f (c β ).

2

Theorem 4.3. If B is a condensed Boolean algebra, I ⊆ B is an ideal and f : B → B / I is a homomorphism such that |Rng( f )| = ℵ1 , then f is a trivial homomorphism. Proof. The set A := {a | f (a) = a/ I } is a subalgebra of B. By Proposition 4.2, | A | = ℵ1 . Hence by Lemma 3.4, there is B ( I ) ⊇ cmpl B ( I ). So f is trivial. 2 a countable secluded ideal I such that A ⊇ C± Corollary 4.4. Let B be a condensed Boolean algebra, I , J be ideals of B, and f : B / I → B / J be a homomorphism. Suppose that |Rng( f )| = ℵ1 . Then there is a countable secluded ideal K of B such that I ∩ cmpl( K ) ⊆ J ∩ cmpl( K ), and for every a ∈ cmpl( K ), f (a/ I ) = a/ J .

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Proof. Let π be the canonical homomorphism from B to B / I and g = f ◦ π . By Theorem 4.3, there is a countable secluded ideal K ⊆ B such that for every a ∈ cmpl( K ), g (a) = a/ J . Let a ∈ I ∩ cmpl( K ). Then

a/ J = g (a) = f









π (a) = f 0 B / I = 0 B / J .

So a ∈ J . We have shown that I ∩ cmpl( K ) ⊆ J ∩ cmpl( K ). Let a ∈ cmpl( K ). Then

f (a/ I ) = f ◦ π (a) = g (a) = a/ J .

2

Corollary 4.5. Let B be a condensed Boolean algebra, I be an ideal of B, and f : B / I → B be a homomorphism. Suppose that |Rng( f )| = ℵ1 . Then there is a countable secluded ideal J of B such that: (1) I ⊆ J . (2) For every a ∈ cmpl( J ), f (a/ I ) = a. Proof. Let π : B → B / I be the canonical homomorphism of B and I and g = f ◦ π . Then g is an endomorphism of B and |Rng( g )| = ℵ1 . So by Theorem 4.3, there is a secluded ideal J of B such that for every a ∈ cmpl( J ), g (a) = a. Suppose by contradiction that I ⊆ J . Then since J is secluded, there is a ∈ I ∩ cmpl( J ) \ {0}. Then g (a) = 0 = a. A contradiction, so I ⊆ J . Let a ∈ cmpl( J ). Then f (a/ I ) = g (a) = a. 2 Recall that a Boolean algebra B is said to be quotient-categorical, if | B | = ℵ1 , and every uncountable quotient of B is isomorphic to B. Definition 4.6. Let I be an ideal in a Boolean algebra B. A subalgebra C of B is a retract of I in B, if I ∪ C generates B and C ∩ I = {0 B }. If every ideal of B has a retract, then B is said to be retractive. (Stated topologically, a Boolean algebra B is retractive iff every closed subset F of Ult( B ) is a retract of Ult( B ), that is, F is the range of a projection of Ult( B ).) We next show that a condensed Boolean algebra is never quotient-categorical. We also observe that condensed Boolean algebras are not retractive. Corollary 4.7. Let B be a condensed Boolean algebra. (a) Let I ⊆ B be an ideal. Then I has a retract iff either I is uncountable (and hence B / I is countable (Corollary 3.8)), or there is a countable secluded ideal J such that I ⊆ J . (b) B is not retractive. (c) B is not quotient-categorical. Proof. (a) We prove ⇒. If I is uncountable, then B / I is countable. This implies that I has a retract. Suppose that J is a countable secluded ideal and I ⊆ J . The algebra C := J ∪ − J is countable, so C , I  has some retract A 1 . Let A = cl B ( A 1 ∪ cmpl( J )). Then A is a retract for  B , I . Proof of prove ⇐. Let I be a countable ideal and A be a retract for I . There is a countable secluded ideal J such that A ⊇ cmpl( J ). It follows that I ⊆ J . (b) Since I 1 ( B ) is not contained in a countable secluded ideal, I 1 ( B ) does not have a retract. (c) Let C = B / I 1 ( B ). We show that C  B. Suppose by contradiction that f : C ∼ = B. Define g : B → B as follows: g (a) = f (a/ I 1 ( B )). Then g is an endomorphism from B onto B, and hence, Rng( g ) is uncountable. It follows that g is a trivial endomorphism. That is, there is a countable secluded ideal I of B such that g (a) = a for every a ∈ cmpl B ( I ). Clearly, At( B ) ∩ cmpl B ( I ) = ∅. Let a ∈ At( B ) ∩ cmpl B ( I ). Then g (a) = a. On the other hand, since a ∈ I 1 ( B ), it follows that g (a) = 0. A contradiction. 2 In [2] we defined the notion of a tightly Hausdorff space. In [3] (in preparation) we shall observe that a retractive space is tightly Hausdorff, and prove assuming ♦ℵ1 , that there is a thin–tall space which is tightly Hausdorff but not retractive. We show here that the Stone space of a condensed Boolean algebra is not tightly Hausdorff. Let X be a topological space and x ∈ X . Then Nbr X (x) denotes the set of open neighborhoods of x. If A is a set of pairwise disjoint subsets of X , then





ac X (A) := x ∈ X for every U ∈ Nbr(x), { A ∈ A | A ∩ U = ∅} is infinite . Definition 4.8. (a) Let X be a topological space and A ⊆ X . A family U := {U a | a ∈ A } is a tight Hausdorff system for A, if for every a ∈ A, U a ∈ Nbr(a), U is pairwise disjoint, and for every B , C ⊆ U : if

{a ∈ A | B ∩ U a = ∅} = {a ∈ A | C ∩ U a = ∅},

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then







1521



ac { B ∩ U a | a ∈ U } = ac {C ∩ U a | a ∈ U } . (b) A space X is tightly Hausdorff space if for every A ⊆ X : if A with its relative topology is a discrete space, then A has a tight Hausdorff system. Corollary 4.9. The Stone space of a condensed Boolean algebra is not tightly Hausdorff. Proof. Let B be a Boolean algebra, A ⊆ B and b ∈ B. We say that b almost does not cut A, if {a ∈ A | a · b, a − b = 0 B } is finite. Let B be a superatomic Boolean algebra. Then the following are equivalent. (1) Ult( B ) is tightly Hausdorff.  ( B ) be such that for every a ∈ A there is a ∼ B a such that for every b ∈ A \ {a}, b − a ∼ B b. Then there is (2) Let A ⊆ At a pairwise disjoint family C := {ca | a ∈ A } such that for every a ∈ A, ca ∼ B a and such that for every b ∈ B, b almost does not cut C .

 1 ( B ) be such that for every Let B be a condensed BA, and assume by contradiction that B is tightly Hausdorff. Let A ⊆ At B  b ∈ At1 ( B ) there is a unique a ∈ A such that a ∼ b. Then by (2), there is a pairwise disjoint family C := {ca | a ∈ A } such that for every a ∈ A, ca ∼ B a and such that for every b ∈ B b almost does not cut C . Let





D = b ∈ B for every c ∈ C , d  c or d · c = 0 B . D is a subalgebra of B and D is uncountable. We show that D does not contain the complement of a countable secluded /D ideal. Let I be a countable secluded ideal. There is c ∈ C such that (At( B )  c ) ∩ I is finite. Let e ∈ (At( B )  c ) \ I . Then e ∈ and e ∈ cmpl( I ). So D  cmpl( I ). 2 We next observe that a condensed Boolean algebra cannot be well-generated. A Boolean algebra B is well-generated, if B has a sublattice L such that L generates B and  L , < B  is well-founded. (See [1].) Corollary 4.10. If B is a condensed Boolean algebra, then B is not well-generated. Proof. We show that B does not have an uncountable well-founded sublattice. Let A ⊆ B be uncountable. We show that the lower semi-lattice generated by A is not well-founded. Either A ∩ I ( B ) or A ∩ − I ( B ) are uncountable. Suppose first that A ∩ − I ( B ) is uncountable. We may then assume that A ⊆ − I ( B ). Wedefine by induction {ai | B i ∈ ω} ⊆ A. Choose a0 ∈ A. Suppose that  ai has been defined for every i < n. The  interval [ i
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5. The Ostaszewski case, packed Boolean algebras Definition 5.1. (a) Let B be unitary. A wide interval of B is a set of the form [c , −e ] B , where c , e ∈ I ( B ) and c · e = 0. We call c , e  a wide interval pair of B. Set Wip( B ) := {c , e  ∈ I ( B ) × I ( B ) | c · e = 0}. (b) Let B be unitary and P be a subset of B. We say that P is somewhere dense, if there is c , e  ∈ Wip( B ) such that for every c 1 , e 1  ∈ Wip( B ): if [c 1 , −e 1 ] B ⊆ [c , −e ] B , then P ∩ [c 1 , −e 1 ] B = ∅. (c) A Boolean algebra B is called a packed Boolean algebra if: (1) B is thin–tall and unitary. (2) Every uncountable subset of I ( B ) is somewhere dense (and therefore every uncountable subset of B is somewhere dense). We show that under ♦ℵ1 packed algebras exist, and that packed algebras are Ostaszewski algebras. Theorem 5.2. Assume ♦ℵ1 . There is a packed Boolean algebra. Proof. In the definition of a PI-system remove the requirement that D is infinite, and in the definition of a condensed PI-system remove the requirement that for every α < ω1 there is d ∈ D M such that rk( B M  d)  α . For a packed BA we need that D M be the empty set. The proof of the Construction Theorem 2.7 can be repeated, starting with a PI-system M ω in which D ω = ∅, and omitting those parts in the proof in which D M is enlarged. One obtains a (much simpler) PI-system M is which B M = B M := B and D M = ∅. The argument which shows that every uncountable subset of I ( B ) is somewhere dense remains the same. The resulting algebra B is packed. 2 Proposition 5.3. Let B be a packed Boolean algebra. (a) Let I ⊆ B be an uncountable ideal. Then there is a ∈ − I ( B ) such that I ⊇ I ( B )  a. (b) B is an Ostaszewski algebra. Proof. (a) Any meet-closed uncountable subset P of I ( B ) contains a wide interval. The proof of this fact is the same as in Proposition 3.7(b). Let I be an uncountable ideal. Then I ∩ I ( B ) is an uncountable ideal. Let [c , −e ] be a wide interval such that I ∩ I ( B ) ⊇ [c , −e ]. Then I ∩ I ( B ) ⊇ I ( B )  (−e − c ) and −e − c ∈ − I ( B ). (b) We prove that B is sub-Ostaszewski. By the remark following Definition 1.7, it suffices to show that if I ⊆ B is an uncountable ideal, then B / I is countable. Let I be such an ideal. By Part (a), there is a ∈ − I ( B ) such that I ⊇ I ( B )  a. Then {b/ I | b ∈ B  (−a)} = B / I . So B / I is countable. To show that B is Ostaszewski it remains to prove that every countable closed subset of Ult− ( B ) is compact. Let F be such a set. Define I = {a ∈ I ( B ) | for every x ∈ F , a ∈ / x}. Note that U IB = Ult− ( B ) \ F . Hence I is uncountable. Let b ∈ − I ( B ) be such that I ( B )  b ⊆ I . Set J = B  −b. We show that F ⊆ U BJ . Let x ∈ F . Suppose by contradiction that b ∈ x. Let c ∈ x ∩ I ( B ). Then c · b ∈ I ( B )  b and hence c · b ∈ I . However, c · b ∈ x. This contradicts the definition of I . So b ∈ / x. Hence −b ∈ x. So x ∈ U BJ . U BJ is compact, U BJ ⊆ Ult− ( B ) and F ⊆ U BJ . So F is compact. 2 As mentioned, an Ostaszewki algebra cannot be downward-categorical. In particular, packed Boolean algebras are not downward-categorical. However, all the other properties of a condensed BA, those which are discussed in Section 4, have a counterpart in the “packed” case. In essence, if in the condensed case we say that something happens “outside a countable secluded ideal”, then in the “packed” case we obtain that the same thing happens “outside a countable principal ideal”. In particular, we have the following conclusion. Corollary 5.4. Let B be packed. Then B is rigid in the following sense. Suppose that I , J are ideals in B and f : B / I → B / J is a homomorphism with an uncountable range. Then there is a principal ideal K of B such that |cmpl( K )|  ℵ0 , I ∩ K ⊆ J ∩ K , and for every a ∈ K , f (a/ I ) = a/ J . Remark 5.5. The following variation of Corollary 5.4 can be proved assuming only (CH). There is a thin–tall BA B such that the following holds. Let I , J be ideals in B with | I |, | J |  ℵ0 , and let f : B / I → B / J be a homomorphism. Then there is a principal ideal K of B such that |cmpl( K )|  ℵ0 , I ∩ K ⊆ J ∩ K , and for every a ∈ K , f (a/ I ) = a/ J . In fact, there are 2ℵ1 pairwise non-isomorphic such algebras. 6. An observation and a question Propositions 6.1 and 6.2 were found by M. Weese [7] 1991.

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Proposition 6.1. Let B be a downward-categorical algebra. Then B is superatomic. Proof. Suppose that B is not superatomic. So B contains a chain isomorphic to Q. Let b ∈ B be such that B  b contains a chain isomorphic to Q and B  −b is uncountable. For each countable ordinal α let C α be a subalgebra of B  b isomorphic to Bα . So C α × ( B  −b) is uncountable and thus isomorphic to B. For every α < ω1 let aα ∈ B be such that B  aα ∼ = Bα , and let C be the subalgebra of B generated by {aα | α < ω1 }. Then C is superatomic and C ∼ = B. A contradiction. 2 Proposition 6.2. Assume 2ℵ0 < 2ℵ1 . Let B be a downward-categorical algebra. Then either B is thin–tall or B is isomorphic to the algebra of finite and cofinite subsets of ω1 .

 α (B) Proof. Suppose that B is not thin–tall. Clearly B is unitary and rk( B )  ω1 . Let α < ω1 be the first ordinal such that At is uncountable. Assume by contradiction that α > 0. So At( B ) and I α ( B ) are countably infinite. Let C be the subalgebra  α ( B ). Since C is uncountable, it is isomorphic to B. generated by I α ( B ) ∪ At  α ( B ), we denote by B ( A ) the subalgebra of B generated by I α ( B ) ∪ A. Clearly, For each uncountable subset A of At  α ( B )}| = 2ℵ1 . That is, B has 2ℵ1 distinct uncountable subalgebras. |{ B ( A ) | A ⊆ At It is trivial that if f , g are embeddings of B into B and f  At( B ) = g  At( B ), then f = g. So B has at most 2ℵ0 embeddings into itself. A contradiction. It follows that α = 0. Hence |At( B )| > ℵ0 . This implies that the Boolean algebra F of finite and cofinite subsets of ω1 is embeddable in B. So B ∼ = F. 2 We conclude with a question. Question 6.3. (a) Is it consistent that there is a thin–tall downward-categorical BA B such that B has an automorphism f  α ( B ) such that f (a) · a = 0? such that for every α < ω1 there is a ∈ At (b) Is it consistent that there is a thin–tall downward-categorical BA B such that B has an automorphism f such that for every for every a ∈ At( B ), f (a) = a? Symbol index C  a = {c ∈ C | c  a} a ∼B b b · A = {b · a | a ∈ A } [c , e ] B := {b ∈ B | c  b  e }. This is defined for c , e ∈ C , where B is a subalgebra of C (c , e ) B , (c , e ] B , [c , e ) B A  B. A and B are Boolean algebras M  N. M and N are PI-systems M G . The extension of M obtained from a directed set G ⊆ P M B 1  B 2 . The subalgebra of B 1 × B 2 generated by I ( B 1 ) × {0 B 2 } ∪ {0 B 1 }× I ( B 2 )

1504 1506 1506 1506 1506 1508 1508 1510 1516

Notation index c iG

1510

p

ci

 ( B ) = {a ∈ I ( B ) | B  a is unitary} At  ( B ) | rk(a) = α }  α ( B ) := {a ∈ At At B± ( I ) := I ∪ − I Bα . The countable superatomic BA with rank  unitary p c p = i ∈ω c i B C± ( I ) := B± (cmpl B ( I ))

cl B (C ). The subalgebra of B generated by C cmpl B ( E ) = {b ∈ B | for every e ∈ E , b · e = 0} Cnvnt( Q , M ; N ) D ( X ). The set of non-isolated points of X D α ( X ). The α ’s Cantor–Bendixon derivative of X e X . The point with highest rank in X g (t , p ) h(t , p ) I ( B ) = {a ∈ B | rk B (a) < rk( B )}

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Iα (B) I M . The ideal generated by I ( B M ) ∪ D M intrvl(t , p ) := [ g (t , p ) , −h(t , p ) ] Isol( X ). The set of isolated points of X P M . The forcing of a countable PI-system M P QM := {c , e  ∈ Wip( M ) | P ∩ [c , −e ] = ∅} rk( X ). Rank of the scattered space X rk X ( F ). Rank of a closed subset with its induced topology rk( B ). Rank of a superatomic Boolean algebra rk B (a). Rank of the element a in B rk( I ). Rank of the ideal I rk( M ). The rank ofM. rk( M ) = rk( B M ) p T b := { p ∈ P | b ⊆ i ∈ω c i } T i ,k,β T i ,d T t, Q M . The set of terms of M U IB = {x ∈ Ult( B ) | x ∩ I = ∅} Ult( B ). The Stone space of B Ult− ( B ) = Ult( B ) \ {e Ult( B ) } Wip( M ) = {c , e  ∈ I ( B ) × I M | c · e = 0} Wip( B ) = {c , e  ∈ I ( B ) × I ( B ) | c · e = 0}

τ

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Definition index Boolean space. 0-dimensional compact Hausdorff space condensed Boolean algebra condensed PI-system dense set of intervals in M downward-categorical Boolean algebra narrow condensed Boolean algebra narrow condensed PI-system nowhere dense set in M Ostaszewski Boolean algebra packed Boolean algebra PI-system pure ideal quotient-categorical Boolean algebra quotient-far rank of a scattered space retract retractive rich Boolean algebra scattered space secluded ideal somewhere dense set in a PI-system somewhere dense set in a unitary BA strongly non-Ostaszewski sub-Ostaszewski Boolean algebra superatomic Boolean algebra thin–tall Boolean algebra thin–tall space trivial endomorphism trivial homomorphism unitary Boolean algebra unitary space well-generated Boolean algebra wide interval of a PI-system wide interval of a unitary BA

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